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,  PRACTICAL 
DESCRIPTIVE    GEOMETRY 


BY 

WILLIAM  GRISWOLD  SMITH,  M.  E. 

ASSISTANT  PROFESSOR   OF   DESCRIPTIVE   GKOMETRY   AND 
KINEMATICS,   ARMOUR   INSTITUTE   OF   TECHNOLOGY 


McGRAW-HILL    BOOK   COMPANY 

239  WEST  39TH  STREET,  NEW  YORK 

6  BOUVERIE  STREET,  LONDON,  E.  G. 

1912 


;     ."•    ' 


COPYRIGHT,  1912,  BY  THE 
McGRAW-HiLL  BOOK  COMPANY 


PRINTED-AND-ELECTROTYPED 
THE.  MAPLE.  PRESS.  YORK.  PA 


PREFACE 

In  presenting  a  new  text-book  on  Descriptive  Geometry,  the 
writer  is  fully  aw^are  of  the  excellence  of  existing  treatises,  and 
appreciates  what  they  have  contributed  toward  a  higher  standard 
of  technical  training.  The  authors  of  these  treatises,  pursuing 
their  several  courses,  have  carried  the  development  of  the  subject 
to  a  high  plane. 

However,  in  spite  of  the  indisputable  excellence  of  many  of  the 
text-books,  they  seem  to  have  failed  to  arouse  the  interest  of 
the  student,  partly  by  ignoring  the  practical  applications,  and 
partly  by  making  only  a  slight  attempt  to  present  the  subject 
attractively.  Some  of  the  books  are  valuable  only  for  reference 
and  are  useless  in  the  class  room;  others  are  incomplete  in  the 
essentials;  some  are  faddish,  emphasizing  certain  features  and 
treating  the  rest  inadequately,  while  even  the  best  convey  to  the 
student  only  a  very  slight  idea  of  the  practical  value  of  the 
subject.  This  has  given  rise  to  the  belief,  prevailing  almost 
universally  in  the  student  body,  that  Descriptive  Geometry  is 
merely  a  disciplinary  study,  having  little  or  no  relation  to  the 
life  work  of  the  individual. 

The  aim  of  the  writer  has  been,  therefore,  to  present  the 
subject  to  the  student  in  a  simple  manner,  as  progressively  as 
possible,  reminding  him  constantly  of  the  relation  which  exists 
between  Descriptive  Geometry  and  Practical  Drafting;  and  to 
avoid  needless  difficulty  by  using  language  and  directions  of 
the  greatest  possible  clarity.  The  writer  believes  that  a  thorough 
knowledge  of  the  subject  is  achieved  not  through  much  study  of 
the  text,  but  by  working  exercises.  To  this  end  he  has  provided 
a  large  number  of  exercises,  scattered  through  the  text,  of  con- 
siderable variety  and  capable  of  infinite  multiplication  by  the 
clever  instructor. 

An  examination  of  the  subject  will  reveal  the  following 
features  and  innovations: 

(1)   A  thorough  drill  in  fundamentals. 


248526 


vi  PREFACE 

(2)  Repetitions  of  statements  for  the  sake  of  emphasis. 

(3)  Notation  comprehensive,  yet  reduced  to  its  lowest  terms. 

(4)  Analyses  separated  from  proofs. 

(5)  Tabulated  order  of  the  operations  in  each  analysis  and 
construction. 

(6)  Exercises  to  the  number  of  860,  of  which  about  one-fourth 
are  such  as  may  be  met  in  actual  practice.     Most  of  these  are 
dimensioned  for  a  space  of  standard  size. 

(7)  Notes  of  exceptions,   checks  on  solutions,  special  cases, 
hints,    and   rules   presented   wherever  they  seem  necessary   or 
advisable. 

(8)  Brief  treatment  only  of  such  special  subjects  as  Shades 
and  Shadows,  Perspective,  etc. 

(9)  The  original  seventeen  " Point,  Line,  and  Plane"  problems 
expanded  to  forty-three. 

(10)  New  treatment  of  surfaces,  less  importance  being  attached 
to  passing  tangent  planes,  and  more  importance  to  plane  sections, 
intersections,    developments,    and   practical    applications.     For 
example,  there  are  about  thirty  exercises  which  are  based  on 
practical  illustrations  of  the  various  warped  surfaces,  showing 
actual  uses  for  all  the  listed  varieties. 

The  expansion  of  the  problems  noted  in  feature  (9)  is  justified, 
the  writer  believes,  in  order  to  fully  cover  the  ground,  and  to 
provide  the  necessary  steps  in  the  development  of  the  subject. 
Some  of  them  are  special  cases  and  some  require  little  or  no 
change  in  the  analysis  from  those  of  preceding  problems. 

Students  undertaking  the  study  of  Descriptive  Geometry 
should  have  a  fair  knowledge  of  Mechanical  Drawing,  at  least  as 
much  as  is  usually  included  in  High  School  courses.  The  min- 
imum prerequisite  should  amount  to  the  work  contained  in  the 
first  eight  chapters  of  French's  "  Engineering  Drawing/'  or 
in  Reid's  "  Mechanical  Drawing"  entire. 

In  preparing  this  text-book  the  writer  has  consulted  most  of 
the  standard  works  on  the  subject,  and  desires  to  make  special 
acknowledgement  for  ideas  and  information  to  the  excellent 
books  by  MacCord,  Church  and  Bartlett,  Phillips  and  Millar, 
and  Randall.  Most  of  the  exercises  are  original,  but  many  have 
been  adapted  from  such  admirable  books  of  exercises  as  those 
by  Professors  Hood,  Marshall,  and  Fishleigh,  as  well  as  the 
problem  files  of  Armour  Institute  of  Technology,  and  the  books 
of  the  American  School  of  Correspondence. 


PREFACE  vii 

The  writer  wishes  to  thank  his  colleagues,  Robert  V.  Perry, 
Henry  L.  Nachman,  John  S.  Reid,  and  Charles  R.  Swineford,  of 
the  Faculty  of  Armour  Institute  of  Technology,  for  valuable 
suggestions  and  encouragement. 

WILLIAM  GRISWOLD  SMITH. 
ARMOUR  INSTITUTE  OF  TECHNOLOGY, 
Chicago,  June  20,  1912. 


J  SMiA 


PRACTICAL  DESCRIPTIVE  GEOMETRY 

CHAPTER  I 

DEFINITIONS,  NOTATION,   PRELIMINARY  THEOREMS,  AND 

EXERCISES. 

1.  Definition. — Descriptive  Geometry  is  the  Science  of  Mechan- 
ical Drawing. 

Mechanical  Drawing  is  the  Art  of  representing  objects  on  paper, 
or  other  flat  surface,  by  means  of  projections  of  the  objects  on 
imaginary  planes. 

Projections  are  images  of  objects  made  by  imaginary  lines 
projected  from  the  objects,  and  are  variously  made  for  various 
uses,  and  may  be  thus  classified: 

Orthographic  Projection:  objects  projected  by  perpendicular 
lines. 

Perspective  or  Scenographic  Projection:  objects  projected  by 
converging  lines. 

Oblique  Projection:  objects  projected  by  parallel  oblique  lines. 

Isometric  Projection,  Cavalier  Projection,  and  Pseudo  Per- 
spective are  variants  on  Perspective,  using  Orthographic  or 
Oblique  Projection.  They  are  termed  collectively  Single  Plane 
Projection,  and  treated  in  a  separate  chapter. 

ORTHOGRAPHIC  PROJECTION 

2.  Discussion. — This   is   by   far  the  most  important  class  of 
projection,  and  is  almost  universally  used  in  making  a  "working 
drawing"  and  in  "machine  sketching."     To  thoroughly  under- 
stand its  significance,  let  us  first  consider  the  difference  between 
a  picture  and  a  working  drawing. 

A  picture,  or  perspective,  is  a  representation  of  an  object, 
group,  or  scene,  as  it  appears  to  the  eye.  Fig.  1  shows  the  way 
a  picture  is  projected.  The  lines  of  light  from  any  object  reach 
the  eye  in  converging  lines.  In  this  figure,  A  is  the  point  of 
sight,  B  and  C  are  objects  of  equal  size,  but  at  different  distances 
from  A,  and  D  is  a  plane  of  projection  between  the  eye  and  the 

1 


PRACTICAL  .DESCRIPTIVE  GEOMETRY 


objects.  Converging  rays  (shown  by  the  dotted  lines),  from  the 
tips  of  B  and  C  to  the  eye,  pierce  the  plane  D  at  certain  points. 
These  points  determine  the  size  of  the  pictures  of  B  and  C,  and 
the  images  B'  and  C'  are  these  pictures.  As  B  is  the  nearer, 
B'  is  the  larger.  Thus  distant  objects  appear  smaller  than 
nearer  ones  of  equal,  and  often  smaller  size.  By  means  of  this 
fact,  and  the  knowledge  of  the  beholder  of  the  comparative 


FIG.  1. 

sizes  of  the  objects,  a  picture  shows  the  entire  composition  of  a 
scene. 

The  working  drawing  takes  no  such  account  of  distances. 
Projections  are  made  by  rays  perpendicular  to  the  plane,  and 
therefore  parallel  to  each  other,  which  means  that  equal-sized 
objects,  whether  near  or  remote,  will  have  equal-sized  projections. 

Fig. .  2  shows  this  fact.     In  orthographic  projection  the  eye  is 


«  ^*  — 

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i 

A  •*»• 

A/2- 

B' 

C1 

-4»  

A    ^8t  

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^ 

^ 

\D 

FIG.  2. 

considered  to  be  at  an  infinite  distance.  If  the  effect  of  a  finite 
position  of  the  eye  is  desired,  the  eye  must  be  shifted  (as  A  in 
Fig.  2)  to  the  line  of  each  ray.  This,  of  course,  is  an  artificial 
condition,  and  proves  that  the  working  drawing  is  not,  and 
cannot  be,  a  correct  visual  image  of  the  object. 

This  fact  would  seem  to  be  a  disadvantage  against  the  use  of 
orthographic  projection,  but  it  is  actually  much  more  useful  in 


DEFINITIONS,  NOTATION,  PRELIMINARY  THEOREMS 

manufacture  and  construction  than  the  perspective,  and  much 
easier  for  the  draftsman  to  use.  On  account  of  the  complexity 
of  many  machines  and  structures,  many  dimensions  are  necessary 
in  many  directions,  interior  dimensions,  thicknesses,  and  irregu- 
larities have  to  be  shown,  which  are  all  but  impossible  in  a 
picture.  This  is  accomplished  by  projecting  the  object  ortho- 
graphically  on  two  or  more  planes,  called  the  Planes  of 
Projection. 

THE  PLANES  OF  PROJECTION 

3.  The  two  principal  planes  are  the  Vertical  and  Horizontal. 
The  Vertical  Plane  (which  we  shall  hereafter  write  V)  is  a  plane 
normal   to   the  earth's   surface;   that  is,   perpendicular  to   the 
apparent  surface  of  the  earth.     The  Horizontal  Plane  (hereafter 
written  H)  is  a  plane  perpendicular  to  any  vertical  line;  that  is, 
it  is  parallel  to  the  apparent  surface  of  the  earth.     H  and  V  are 
perpendicular  to  each  other.     The  projections  on  these  two  planes 
determine  the  relative  position  of  any  two  or  more  points  in 
space.     There  are  conditions  arising  often,  which  render  a  third 
plane  desirable,  and  indeed  necessary,  in  the  solution  of  certain 
problems.     This  third  plane  is  perpendicular  to  both  H  and  V, 
and  is  called  the  Profile  Plane,  which  will  always  be  written  P. 

Any  number  of  auxiliary  planes,  which  may  be  parallel,  per- 
pendicular, or  oblique  to  H  or  V,  may  be  used,  according  to  the 
complexities  of  the  problem.  These  are  commonly  called  Q  in 
this  book.  (In  the  architect's  plans  of  a  big  building,  two  differ- 
ent vertical  projections  may  be  needed,  two  profiles,  a  horizontal 
projection  for  every  floor,  basement,  and  roof,  auxiliary  pro- 
jections of  certain  oblique  features,  sectional  interior  projections, 
and  a  hundred  " detail"  projections  of  doors,  windows,  stair- 
cases, elevators,  etc.,  all  in  orthographic  projection.  Besides 
these,  it  is  customary  to  make  a  perspective  of  the  projected 
building  with  colors,  trees,  grass,  walks,  etc.  The  average 
purchaser  cannot  comprehend  working  drawings,  while  the 
picture  is  of  little  use  to  the  workman.  Perspectives  are  for  the 
customers,  orthographic  projections  for  the  workmen.) 

REPRESENTATION  ON  THE  FLAT 

4.  It  is  obvious  that  it  would  be  awkward  and  impracticable 
to  have  a  drawing  board  made  with  horizontal,  vertical,  and 


4  PRACTICAL  DESCRIPTIVE  GEOMETRY 

profile  sides,  and  the  respective  projections  made  on  them. 
Not  only  would  it  be  inconvenient,  but  it  is  entirely  unnecessary. 
Fig.  3  shows  H  and  V  intersecting,  as  they  must,  in  a  line,  called 
the  Ground  Line  (always  designated  GL).  By  revolving  V 
about  GL  as  an  axis,  or  hinge,  it  will  be  brought  into  coincidence 
with  H,  and  H  and  V  projections  can  be  drawn  on  one  surface. 
Similarly,  P  may  be  revolved  into  the  same  surface  about  its 
intersection  with  V  or  H.  (For  mere  solution  of  problems  it  is 


FIG.  3. 

immaterial  whether  P  is  revolved  about  its  intersection,  or 
trace,  with  V  or  with  H,  but  in  practical  work  it  is  far  more 
often,  if  not  always,  revolved  about  the  F-trace.)  In  this  work 
the  profile  will  always  be  represented  as  rotated  about  its  F-trace. 

Note. — When  working  at  the  blackboard,  it  will  be  easier  to  regard  the 
foregoing  rotation  as  being  made  thus:  rotate  H  into  coincidence  with  V, 
by  dropping  the  front  portion  of  H,  and  raising  the  part  behind  V.  Which- 
ever way  it  is  considered,  the  First  and  Third  angles  are  opened,  and  the 
Second  and  Fourth  Angles  are  closed. 

THE  FOUR  DIHEDRAL  ANGLES 

5.  It  will  be  seen  from  Fig.  3  that  the  planes  of  projection  do 
not  end  at  GL.  In  fact,  they  are  unlimited  in  extent,  and 
divide  space  into  four  dihedral  angles,  with  GL  as  their  common 
intersection.  In  folding  into  the  flat,  therefore,  no  portion  of 
the  drawing  board  is  monopolized  by  either  H  or  V.  There  is 
a  portion  of  H  above  GL,  and  a  portion  below  GL.  There  is  a 
portion  of  V  above  GL,  and  a  portion  below  GL.  (This  circum- 
stance will  be  found  essential  to  the  solution  of  certain  problems 
of  abstract  character.  In  practical  work  the  different  projec- 
tions, or  views,  as  they  are  often  called,  are  entirely  separated 
from  each  other,  and  definite  spaces  are  reserved  for  H,  V.. 
and  P.) 


DEFINITIONS,  NOTATION,  PRELIMINARY  THEOREMS  5 

The  four  dihedral  angles,  in  any  of  which  objects  may  be  imagined,  are 
numbered,  as  shown  by  the  Roman  numerals  in  Fig.  3. 
First  Angle,  in  front  of  V,  above  H; 
Second  Angle,  behind  V,  above  H; 
Third  Angle,  behind  V,  below  H; 
Fourth  Angle,  in  front  of  V,  below  H. 

The  first  angle  (designated  in  this  work  as  7)  is  chiefly  used  in 
our  problems,  on  account  of  its  accessibility  and  the  ease  of 
presentation  by  the 'teacher  to  the  class,  when  giving  a  solution 
of  a  problem  in  space. 

The  third  angle  (///)  is  commonly  used  in  practical  work, 
for  the  reason  that  the  //-projection  is  usually  made  looking 
down  on  the  object,  and  besides  for  this  angle  the  profile  comes 
out  right-handed  and  detached,  whereas  the  usual  method  of 
swinging  a  profile  from  a  first-angle-projection  will  either  bring 
the  object  out  left-handed,  or  mixed  up  with  the  other  views. 
Many  of  our  practical  problems  are  given  in  ///. 

6.  Commercial  Terminology  of  Projections 

As  the  terms  F-projection,  //-projections,  etc.,  are  not  descrip- 
tive to  non-professionals,  certain  terms  are  used  by  draftsmen. 

The  F-projection  is  often  called  the  Elevation,  Front  View, 
Rear  View,  or  Sectional  Elevation,  as  the  case  may  be. 

The  //-projection  is  called  the  Plan,  Top  View,  Bottom  View, 
or  Sectional  Plan,  as  the  case  may  be. 

The  P-projection  is  called  the  Profile,  End  View,  Right  or 
Left  End  View,  or  Profile  Section,  as  the  case  may  be. 

THE  REPRESENTATION  OF  POINTS,  LINES,  AND  PLANES 

7.  Points. — It  is  obvious  that  the  projection  of  a  point  on  one 
plane  does  not  determine  that  point.     For  example,  let  a  point 
on  the  floor  be  the  projection  of  some  point  in  the  room.     How 
high  is  that  point?     It  may  be  any  distance.     The  point  may  be 
the  projection  itself,  or  the  point  may  be  on  the  ceiling  above, 
or  anywhere  in  between.     To  locate  the**point  in  space  by  one 
projection,  it  is  necessary  to  add  its  distance,  positive  or  negative, 
from  the  projection.     A  better  way  (the  draftsman's  way)  is  to 
locate  the  point  graphically.     By  projecting  that  point  on  the 
wall,  its  height  is  determined  and  shown. 

This  definitely  locates  a  point,  for,  if  perpendicular  lines  be 
erected  from  each  projection  of  a  point,  these  perpendiculars 
will  pass  through  the  point.  As  there  can  be  but  one  intersection 


6 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


of  the  perpendiculars,  the  point  is  absolutely  located  in  space  by 
its  projection  on  two  planes.  The  perpendicular  lines  from  the 
point  to  the  planes  of  projection  are  called  projectors. 

8.  The  H-  and  V-projections  of  a  point  must  lie  on  a  common 
perpendicular  to  GL. 

For  the  //"-projector  and  the  F-projector  are  both  perpendicular 
to  the  respective  planes  of  projection,  thus  determining  a  plane 
perpendicular  to  both  H  and  V,  therefore  perpendicular  to  their 
intersection,  GL.  This  plane  being  perpendicular  to  GL,  its 
intersections  with  H  and  V  will  both  be  perpendicular  to  GL, 
and  at  the  same  point,  which  proves  the  statement.  The  P- 
projection  is  obtained  by  the  intersection  of  the  P-projectors 
from  the  H-  and  F-projections  of  the  point,  when  P  is  revolved 
into  V. 


FIG.  4. 


9.  In  Fig.  4  the  point  A  in  space  is  projected  on  V  in  the  point 
a',  and  on  H  in  the  point  a.  The  projector  Aa'  is  parallel  and 
equal  to  ax  (its  projection  on  H),  because  Aa'  is  parallel  to  H, 
being  perpendicular  to  V.  Likewise  Aa  is  parallel  and  equal  to 
a'x.  This  fact  proves  the  following  theorem. 

Theorem  I. — The  distance  that  any  point  is  from  H  is  equal  to 
the  distance  from  its  V-projection  to  GL;  the  distance  that  any 
point  is  from  V  is  equal  to  the  distance  from  its  H -projection  to 
GL. 

(Note. — The  theorems  in  this  chapter  can  all  be  proved  by  Plane  and 
Solid  Geometry,  and  form  the  basis  of  the  problems  that  follow.  The 
student  should  be  familiar  with  them  and  their  proof.) 


DEFINITIONS,  NOTATION,  PRELIMINARY  THEOREMS  7 

PROJECTIONS  OF  POINTS  IN  THE  VARIOUS  ANGLES 

10.  From  Fig.  4  it  may  be  seen  that,  in  folding  the  planes  into 
the  flat,  a'  will  be  above  GL,and  a  will  be  below  GL. 

This  relation  between  the  projections  only  obtains  in  angle  I. 
Fig.  5  gives  a  pictorial  representation  of  points  in  all  the  angles, 
and  their  revolution  into  the  flat.  A,  in  /,  shows  its  //-projection 
rotated  below  GL.  B,  in  //,  shows  its  //-projection  rotated 


id1 


FIG.  5. 


FIG.  5a. 


above  GL.  C,  in  III,  shows  its  //-projection  rotated  above 
GL.  D,  in  IV,  shows  its  //-projection  rotated  below  GL.  Fig. 
oo,  shows  the  flat  representation  of  A,  B,  C,  and  D. 

Theorem  II. — Points  located  in  front  of  V  have  their  H -pro- 
jections below  GL.  Points  behind  V  have  their  H -projections 
above  GL.  Points  above  H  have  their  V-projections  above  GL. 
Points  below  H  have  their  V-projections  below  GL. 

Theorem  III. — Points  in  H  have  their  V-projections  in  GL, 
and  points  in  V  have  their  H -projections  in  GL. 

Proof. — A  point  in  H  is  zero  distance  from  //,  hence  its 
F-projection  is  zero  distance  from  GL.  (Theorem  I.) 

11.  Designation. — The  designation  of  an  actual  point  in  space 
is  a  capital  letter,  while  the  projections  of  the  point  are  desig- 
nated by  small  (lower  case)  letters.  The  //-projection  of  a 
point  is  distinguished  from  the  F-projection  of  the  same  point 
by  the  use  of  the  "prime"  in  connection  with  the  F-projection. 
Thus,  the  point  A  will  have  its  //-projection  a,  and  its  F-pro- 
jection  a'.  For  the  P-projection  we  use  a  superscript  (p),  ap. 


8 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Revolved  Points. — When  points  are  moved  from  their  original 
position  by  revolution,  or  otherwise,  the  new  position  is  written 
with  a  subscript;  thus,  at  means  that  A  is  in  a  revolved  position 
and  this  is  its  new  horizontal  projection.  The  second  new  posi- 
tion would  be  written  an  and  a'tl  for  the  two  projections,  or  if  it 
were  the  P-projection,  aplt,  etc. 

12.  Notation. — In  giving  out  a  point  for  a  problem,  we  write; 
(1)  its  distance  to  the  right,  measured  along  GL,  (2)  the  distance 
above  or  below  GL  of  its  F-projection,  and  (3)  the  distance  above 
or  below  GL  of  its  //-projection.  Fig.  6  shows  the  location  on 


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V 

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li^jdV     *i 

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FIG.  6. 


paper  of  the  following  points,  using  the  notation  that  is  used 
throughout  the  work:  A(f  +  l-l),  B(l£  +  l+i),  0(2^-1+1), 
D(3-i-l),.E(3|,  0-1),  F(44  +  l,  0),  G(5J,  0,  0). 

Example. — The  point  A  is  located  f  in.  to  the  right,  with  its 
F-projection  1  in.  above  GL,  and  the  ^-projection  1  in.  below 
GL.  This  reads  A(j  +  l  — 1).  Plus  always  means  above  GL, 
and  minus  always  means  below  GL,  and  the  F-projection  is 
always  given  before  the  //-projection. 

Note. — The  unit  in  this  notation  may  be  1  inch  for  drawing-board  work, 
and  5  inches  for  blackboard.  Referring  to  Fig.  6,  it  is  not  necessary  to 
put  in  the  dimensions  shown  there,  but  they  are  giveji  the  student,  so  that 
he  may  compare  their  distances  with  the  figures  in  the  notation.  As  no 
work  in  Descriptive  Geometry  can  be  done  without  notation,  the  student 
must  have  a  thorough  drill  in  the  following  exercises: 


EXERCISES  IN  POINT  REPRESENTATION 

Hints. — In  laying  out  any  problem  or  exercise,  do  not  try 
to  locate  the  points  in  space.  No  draftsman  does  that,  he 
does  his  work  on  paper.  Locate  the  points  on  the  paper, 
remembering  that  plus  means  above,  and  minus  means  below 


DEFINITIONS,   NOTATION,  PRELIMINARY  THEOREMT        9 

GL,  and  that  the  first  amount  means  the  distance  to  the  right, 
the  second  the  F-projection,  and  the  third  the  ^-projection. 
The  Roman  numerals  that  appear  in  Fig.  6  refer  to  the  space 
angle  in  which  the  point  is  located. 

LAYOUT  FOR  DRAWING  PLATES 

13.  All  exercises  in  this  work  are  intended  to  fit  in  a  space 
6  in.  X  Tin.,  except  in  a  few  cases,  where  dimensions  are  specified. 
Fig.  7  shows  an  excellent  layout  for  drawing  plates.  A  sheet 


A 

f 

V 

<  6"  ----> 

-  v 

NAME.  . 

« 

p/O  " 

xj 

re/  " 

FIG.  7. 

15  in.  -X  20  in.  is  given  a  1^-in.  margin  on  the  left  for  binding, 
and  a  ^-in.  margin  on  the  other  three  sides.  This  gives  a  work- 
ing space  14  in.  X  18  in.,  divided  into  six  rectangles,  each  6  in.  X 
7  in.,  with  GL  through  the  middle  of  each. 

Inking  Plates. — It  is  not  advisable  to  ink  in  plates,  except 
possibly  the  border  lines,  ground  lines,  and  lettering.  Good 
pencil  work,  and  neat,  legible  lettering  should  be  insisted  on. 

Projectors. — Make  the  lines  running  from  the  projections  to 
GL,  called  projectors,  light,  dotted  lines.  See  Fig.  5a  and  Fig. 
6. 

Letter  all  points. 

EXERCISES 

14.  Locate  on  the  flat  and  designate  in  space  (see  the  Roman  Numerals, 
etc.,  in  Fig.  6)  the  following  points: 
•    1.  A(|  +  2-2),    B(1J— l+2i),    C(2i-2,    0),    D(3,   0,   0),   E(3f-2  +  l), 

F(4i  +  l  +  l),  G(5i,  0-i). 

2.  K(f-l-l),   L(li,   0  +  1),   M(2i,   0-2),   N(3  +  l£,   0),   O(3f-2,   0), 
P(4J  +  1  +  1 


10  PRACTICAL  DESCRIPTIVE  GEOMETRY 

3.  A(|,  1  before  7,  2  below  H),  B(14,  2±  above  H,  I  before  7),  C(2i, 

1  below  H,  1  behind  V),'D(3,  14  behind  7,  14  above  77),  E(3f,  in  H, 

2  behind  V),  F(44,  before  7,  in  70,  G  (si,  in  7,  1  below  #). 

4.  K(f,   in   GL),    L(l$,  H  behind  7,  H  above  #),  M(2|,  in  7,  1|  above 
70,   N(3,   f  below  H,  1  behind  7),  O(3f,  4  before  7,  14  below  #), 
P(44,  in  H,  li  before  7),  and  Q(5J,  1  before  7,  14  above  77). 

Locate  on  the  flat  the  following  points: 

5.  A  (in  ///,  1  to  right,  f  from  7,  2  from  H), 
B(in  II,  2%  to  right,  1  from  H,  1  from  7), 
C(in  /,  4  to  right,  1|  from  7,  f  from  77). 

6.  D(in  H,  14  behind  7,  14  to  right), 

E(in  ///,  3  to  right,  1  from  7,  4  from  H), 
F((in  7,  1  above  77,  44  to  right). 

7.  G  (in  7,  14  to  right,  1  from  V,  1  from  H). 
H(in  7/7,  3  to  right,  2  from  7,  H  from  70, 
K(between  77  and  777,  44  to  right,  1  from  7). 

8.  M(in  77,  14  to  right,  4  from  H,  4  from  77), 
N(between  777  and  77,  3  to  right,  2  from  H), 
O(in  777,  44  to  right,  2  from  7,  1  from  H). 

LINES 

15.  Until  the  subject  of  curved  lines  is  studied,  the  word 
"line"  will  be  taken  to  mean  a  straight  line.  Two  points  deter- 
mine a  line,  and  any  line  that  passes  through  two  points,  con- 
tains those  points.  Further,  the  projections  of  that  line  will 
contain  the  projections  of  those  points. 

Theorem  IV. — Two  projections  of  a  line  will,  in  general,  deter- 
mine tire  position  of  that  line  in  space. 

Proof. — Pass  a  plane  through  the  F-projection  of  some  line, 
and  make  it  perpendicular  to  V.  Also  pass  a  plane  through 
the  77-projection  of  the  same  line,  and  make  the  plane  perpen- 
dicular to  H.  These  planes  will  intersect  in  a  line,  and  that  line 
must  be  the  given  line. 

Theorem  V. — If  a  line  is  parallel  to  H,  its  H-projection  is  par- 
allel and  equal  to  the  line,  and  its  V-projection  is  parallel  to  GL. 

Prove  This. — What  would  be  the  statement  if  the  line  were 
parallel  to  V? 

Theorem  VI. — If  a  line  is  parallel  to  H  and  V,  both  of  its  pro- 
jections will  be  parallel  to  GL. 


DEFINITIONS,  NOTATION,  PRELIMINARY  THEOREMS         11 

Theorem  VII. — If  a  line  is  perpendicular  to  H,  its  H -projection 
will  be  a  point,  and  its  V -projection  will  be  perpendicular  to  GL. 

What  would  be  the  statement  for  a  line  perpendicular  to  V? 

Theorem  VIII. — If  a  line  is  oblique  to  both  H  and  V,  both 
projections  will  be  oblique  to  GL,  except  when  the  line  lies  in  a 
profile  plane. 

16.  Note. — An  oblique  line,  lying  in  a  profile  plane,   must  be  limited 
by  two  points;  otherwise  it  is  indeterminate.     This  is  so,  because  both 
projections  are  perpendicular  to  GL,  and  therefore  lie  in  the  same-  pro- 
jecting plane,   hence   cannot  be   determined  by  the  intersection   of  two 
projecting  planes  (see  Theorem  III).     The  projections  of  any  two  points 
on  the  line  determine  the  line. 

PROJECTIONS  ON  THE  PROFILE  PLANE 

17.  When  a  line  is  so  situated  that  its  H-  and  "F-projections 
are  both  perpendicular  to  GL,  its  true  relation  to  H  and  V,  as 
well  as  to  other  lines  like  itself,  can  only  be  shown  by  the  Profile. 
This  plane,  called  P,  may  be  taken  anywhere,  either  to  right  or 


rh1    r 

r 

I 

*~' 

i 
i 

W^- 

i 
i 

6 

Hfl          V 

j 

\ 

"(oi^  I 

\ 

N 

m                        ^m" 

—  J>-* 

Three  Projections 
of  a  Point.       I 

V 

i 

FIG.  8.                                         FIG. 

So. 

left,  but  is  usually  placed  on  the  right  of  the  problem,  allowing 
sufficient  space  for  its  revolution  into  V  in  either  direction.  Its 
F-trace  is  taken  for  the  axis. 

When  P  is  revolved  into  V,  the  effect,  or  view,  is  the  same  as 
when  the  observer  looks  at  the  plane  in  the  direction  of  GL. 
Fig.  8  shows  H  and  V  in  their  perpendicular  relation,  viewed 
from  the  right  side,  with  the  four  space  angles  in  their  relation 
to  H  and  V.  From  this  it  is  evident  that  in  rotating  the  profile, 


12 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


the  T-trace  of  P  becomes  the  projection  of  V  itself,  and  that 
GL  becomes  the  projection  of  H.  The  profile  of  any  point  is 
then  shown  in  space  its  respective  distances  from  H  and  V. 

Fig.  8  (a)  shows  the  three  projections  of  a  point  M,  f  in.  above 
H,  and  f  in.  in  front  of  V.  The  P-projection  is  obtained  by 
projecting  m'  and  m  to  x  and  y  on  the  respective  traces  of  P, 
and  then  rotating  P  about  its  7-trace,  as  in  the  figure.  M  shows 
in  the  P-projection  that  it  is  f  in.  above  H,  and  f  in.  in  front 
(to  the  left)  of  V. 

If  the  point  lies  in  any  other  angle,  the  revolution  must  be 
made  in  accordance  with  its  location.  The  F-projection  will 
always  move  in  a  line  parallel  to  GL,  whether  it  is  above  or 
below  H.  The  profile  will  be  that  much  above  or  below  H. 
The  //-projection  will  revolve  the  same  as  in  Fig.  8a  if  the 
point  is  in  front  of  V;  that  is,  if  the  //-projection  is  below  GL, 
in  either  angles  /  or  IV.  If  it  is  in  //  or  ///,  its  /f-projection 
is  above  GL,  and  the  profile  rotation  will  be  clockwise,  and  be 
projected  from  GL  to  the  projector  from  its  F-projection. 

Note. — The  profile  is  seldom  necessary  to  the  solution  of  problems.  It 
is  chiefly  used  in  problems  where  there  are  lines  whose  projections  are 
perpendicular  to  GL,  and  planes,  whose  traces  are  parallel  to  GL. 


I 


FIG.  9. 


FIG.  9a. — Flat  projection. 


18.  Fig.  9  shows  a  line,  AB,  in  a  profile  plane,  and  its  projections 
on  H  and  V,  in  picture  form.  Fig.  9  (a)  shows  its  flat  repre- 
sentation, three  projections  being  given.  It  is  obvious  that  the 
profile  is  necessary  for  an  adequate  conception  of  the  line. 

Theorem  IX. — If  a  line  lies  in  V,  its  V-projection  is  the  line 
tself,  and  the  H-projection  is  in  GL. 


Why?     What  is  the  statement  when  the  line  lies  in  HI 


DEFINITIONS,  NOTATION,  PRELIMINARY  THEOREMS         13 


Theorem  X.  If  a  point  or  line  line  lies  in  P,  its  V-projection 
will  be  in  the  V-trace  of  P,  and  its  H-projection  will  be  in  the 
H-trace  of  P. 


19. 


EXERCISES  IN  REPRESENTATION  OF  LINES 


9.  Draw  the  line  A(5—  1  +3)  B(2J  +  1-3),  and  tell  where  it  travels. 
10.  Draw  and  designate  the  lines:  A(|  +  1  -f)  B(2  +  $-!£),  C(2i  +  2  +  l) 


11.  Draw  and  designate  :G(1  -1  +  1)  K(2~i  +  i),  M(2£,  0  +  1)  N(3f,  0  +  2), 
0(4*  +  1,  0)P(5i-l,  0). 

12.  What   kind  of    lines   are   these?     Draw  and   designate:   A(l+l  —  J) 

li),     C(2  +  f-l)     D(31  +  f  +  l),    E(4  +  2-|)    F(4  +  i-l*), 


13.  What   kind   of    lines   are  these?     Draw  and   designate:   A(f  +  l  —  1) 

B(f  +  2-l), 


14.  Tell    what    angles    these    lines    traverse:    P(£+2  +  l)    O(l$  —  1  — 


15.  Tell    what    angles    these    lines  traverse:    A(i  +  l  +  l) 
C(2i-i-li)  D(3i  +  li  +  i),  E(4i  +  l-l)  F(5|-2  +  l). 

16.  From  A(l  +  l  —  1)  draw  a  line  2  in.  long,  parallel  to  V  and  inclined 
45°  to  H.     Does  it  touch  H?     Does  it  touch  V? 

17.  From  B(2  +  l$  —  1J)  draw  a  line  parallel  to  H  and  inclined  30°  to  V. 
How  far  on  this  line  is  it  to  V? 

18.  Draw  the  H-,  F-andP-projections  of  the  lines  A(l  +2-  1)  B(l  +£  —  !£) 
and  C(2i  +  !§  +  !£)  D(2J  +  1  +  £)  .     Locate  P  at  4i  in. 

19.  From  K(lf  —  2  —  1)  draw  a  line  parallel  to  V,  and  45°  to  H,  4  in.  long, 
upward  toward  the  right.     In  what  angle  does  it  terminate? 

20.  From  M  (2  -  2  -  1)  draw  a  line  parallel  to  H,  30°  to  V,  3  in.  long,  toward 
V  to  the  right.     In  what  angle  does  it  terminate? 

REPRESENTATION  OF  PLANES. 

20.   A  plane  other  than  H,  V  and  P  is  shown  by  its  traces; 
that  is,  its  intersections  with  H,  V  and  P.     Usually  the  H  and 


FIG.  10. 


FIG.  10a. 


V  traces  are  sufficient,  and  the  F-trace  is  seldom  used.  Fig.  10 
shows  an  oblique  plane,  t'Tt  (spoken  of  as  the  plane  T),  which 
cuts  V  in  the  line  Tt' ,  its  V-trace,  and  H  in  the  line  Tt,  its 


14  PRACTICAL  DESCRIPTIVE  GEOMETRY 

H-trace.  Fig.  10a  shows  the  flat  representation  of  the  same 
plane.  Tt'  is  one  line  of  the  plane  T,  and  Tt  is  another  line  of 
the  same  plane,  and  the  plane  is  completely  determined  by  them. 

Theorem  XI. — The  two  traces  of  a  plane  completely  determine 
the  plane. 

Note. — The  student  must  on  no  account  think  that  Tt  and  Tt'  are  pro- 
jections of  the  same  line.  Tt  lies  wholly  in  H,  and  its  F-projection  is  in 
GL;  hence,  if  a  point  is  assumed  in  Tt,  its  V -projection  is  in  GL.  Tt'  lies 
wholly  in  V,  and  its  //-projection  is  in  GL;  hence,  if  a  point  is  assumed 
in  Tt',  its  H -projection  is  in  GL. 

21.  Theorem  XII. — The  two  traces  of  any  plane  must  both 
be  parallel  to  GL,  or  must  both  intersect  GL,  and  in  one  point. 

For  GL,  being  common  to  H  and  V,  must  either  intersect  or 
be  parallel  to  all  lines  in  H  and  V.  Tt  is  in  H,  and  Tt'  is  in  V, 
therefore  they  must  be  parallel  to  GL  or  intersect  it,  and  the 
intersections  can  only  be  in  one  point,  because  GL  can  only 
pierce  T  in  one  point.'  If  the  plane  is  parallel  to  GL,  both 
traces  must  be  parallel  to  GL.  Why? 

Note. — The  traces  do  not  end  at  GL,  but  are  indefinite  in  length.  The 
plane  T  occupies  part  of  every  space  angle. 

Theorem  XIII.— If  a  plane  is  parallel  to  V,  it  has  only  one 
trace,  the  H-trace  parallel  to  GL.  What  would  be  true  of  a 
plane  parallel  to  H? 

Theorem  XIV. — A  profile  plane  is  a  plane  perpendicular  to  N 
and  V,  and  its  traces  are  both  perpendicular  to  GL. 

Theorem  XV. — A  plane  perpendicular  to  V,  but  oblique  to  H, 
will  have  its  H-trace  perpendicular  to  GL.  For  example,  see 
the  lid  of  a  box,  as  it  takes  different  positions.  What  would  be 
true  of  a  plane  perpendicular  to  H?  See  a  swinging  door. 

Theorem  XVI.— If  a  plane  passes  through  GL,  it  can  only  be 
satisfactorily  shown  by  its  P-trace.  Both  H-  and  7-traces  are 
in  GL,  and  therefore  the  plane  must  have  either  a  profile  trace 
or  a  point  in  space  to  determine  it.  All  points  in  such  a  plane 
are  projected  on  P  in  the  P-trace  of  that  plane.  Why? 


DEFINITIONS,  NOTATION,  PRELIMINARY  THEOREMS         15 

Theorem  XVII. — Two  parallel  planes  will  have  parallel  traces 
on  any  plane  of  projection. 

Give   a  reason   for  this  from  Solid   Geometry.     Is   the  con-  , 
verse  of  this  always  true?     Take  two  intersecting  planes   par- 
allel to  GL. 

Theorem  XVIII. — Two  planes  perpendicular  to  each  other  do 
not  in  general  have  their  traces  perpendicular.  The  exceptions 
are,  (1)  a  profile  plane  and  a  plane  parallel  to  GL;  and  (2)  two 
perpendicular  planes  that  are  also  perpendicular  to  H,  will 
have  their  //-traces  perpendicular. 

Note. — Planes  perpendicular  to  H  or  V  are  called  projecting  planes, 
because  they  project  every  point  and  line  in  themselves  on  their  traces 
on  the  plane  to  which  they  are  perpendicular. 

NOTATION  OF  PLANES 

22.  As  the  planes  are  represented  by  their  traces,  it  is  suffi- 
cient to  give  the  notation  of  the  traces  to  determine  the  plane. 

Take  the  plane  T  in  Fig.  10,  a  plane  oblique  to  GL,  and  there- 
fore to  H  and  V.  Its  traces  are  t'T  and  Tt,  three  different  points, 
t',  T  and  t,  respectively  in  V,  GL  and  H.  It  will  be  noticed 
that  t'  and  t  are  independent  points,  and  would  be  designated 
(except  for  convenience)  by  different  letters  as  points  in  space. 
The  plane  would  be  written  in  full  t'(l  +2,  0)T(5,  0,  0)t(l,  0-1). 
But,  by  omitting  the  zeros  and  the  repetition  of  letters,  we  can 
reduce  it  to  simpler  terms,  thus: 

T(l +2)5(1-1) 

Rule. — In  the  notation  of  a  plane,  the  F-projection  of  a  point 
in  the  F-trace  is  given  first,  the  point  where  the  traces  intersect. 
GL  second  and  unbracketed,  and  the  //-projection  of  a  point 
on  the  //-trace  third. 

When  a  plane  is  parallel  to  GL  it  has  no  intersection  with  GL; 
that  is,  its  intersection  is  at  infinity,  and  it  is  then  written 
T(  +  2)  oo  (—  1),  the  infinity  sign  indicating  that  the  traces  are 
parallel  to  GL,  and  the  V  and  //-traces  2  in.  above  and  1  in. 
below  GL  respectively.  Both  could  be  above  or  both  below  GL, 
or  the  signs  could  be  opposite  to  those  above. 


16  PRACTICAL  DESCRIPTIVE  GEOMETRY 

Planes  oblique  to  GL,  but  not  intersecting  within  the  problem 
boundaries,  have  a  special  notation,  thus: 


It  can  be  easily  seen  that  this  is  giving  the  notation  of  two 
points  on  each  trace. 

Note. — For  points  and  lines  we  employ  the  first  letters  of  the  alphabet, 
and  for  planes  the  latter  letters. 

EXERCISES  IN  REPRESENTATION  OF  PLANES 

23.  Draw  the  traces  of  the  following  planes,  and  state  their 
relations  to  H,  V  and  P. 

Use  a  dot-and-dash  line  for  all  traces. 

21.  T(l+2i)3(3-2i), 

22.  Q(l+2i)3(l  +  l 

23.  S(  +  li)co(-2),  T(5-l)3(l  +  l). 

24.  U(l+2)(5  +  3)(l+l)(5  +  l^),  X( 

25.  Draw  the  traces  of  the  H  and  F  projecting  planes  of  the  line  A(l  +2  —  £) 
B(3  +  l+li),  and  the  line  C(4-li  +  l$),.D(5i-i-i). 

26.  Show    the    P-traces    of    the   planes  Q(  +  2)oo(  +  l),  R(-li)co(  +  l) 
S(  +  l£)oo(-i),  and  T(  — 2i)oo(-l). 

27. 'Find  the  P-trace  of  a  pl(ane  T  through  GL  and  A(1  +  1J  —  I),  and 
the  P-trace  of  S  through  GL  and  B(2J  — 1  — $). 

Note. — To  find  the  P-trace  of  any  plane,  take  the  plane  P  at  any  convenient 
place,  where  it  will  intersect  the  H  and  F-traces.  Then  treat  the  line 
connecting  these  intersections  (the  P-trace)  as  you  would  any  other  profile 
line,  and  revolve  it  as  in  Art.  17. 

THEOREMS  RELATING  TO  LINES  AND  PLANES 

24.  Theorem  XIX. — Two  lines  parallel  in  space  have  parallel 
projections  on  any  plane. 

Theorem  XX. — Two  lines  perpendicular  in  space  do  not 
ordinarily  have  their  projections  perpendicular. 

Special  Case. — If  one  of  these  lines  is  parallel  to  H,  the  H-pro- 
jections  of  the  two  lines  will  be  perpendicular. 

Fig.  11  shows  a  line  AB,  parallel  to  H,  and  BC  (any  one  of 
four  different  lines),  whose  H  -projection  is  perpendicular  to  ab. 
Any  one  of  the  lines  BC  will  be  perpendicular  to  AB. 


DEFINITIONS,  NOTATION,  PRELIMINARY  THEOREMS         17 

Proof. — If  AB  is  parallel  to  H,  any  plane  perpendicular  to 
AB  will  be  perpendicular  to  H ,  and  will  therefore  be  a  projecting 
plane  on  H  of  all  lines  contained  in  it.  Let  the  student  finish 
the  reasoning  to  show  that  all  lines,  whose  //-projections  are 
be,  will  be  perpendicular  to  AB. 


FIG.  11. 


25.  Theorem  XXI. — Two  intersecting  lines  in  space  must  have 
their  H  and  V-projections  intersect  in  points  in  a  line  perpen- 
dicular to  GL.  Prove. 


Theorem  XXII. — Two  intersecting  lines,  both  parallel  to  H,  are 
projected  on  H  in  lines  parallel  and  equal  to  themselves,  and 
the  included  angle  will  be  equal  in  projection  to  the  angle  in 
space.  Prove. 

Theorem  XXIII. — A  line,  that  is  parallel  to  an  oblique  plane, 
is  not,  in  general,  projected  parallel  to  the  traces.  The  exception 
is  only  possible  when  the  traces  are  both  parallel  or  both  per- 
pendicular to  GL. 

Theorem  XXIV. — A  line,  that  is  perpendicular  to  a  plane, 
always  has  its  projections  perpendicular  to  the  respective  traces 
of  the  plane. 

In  Fig.  12  let  BC  be  perpendicular  to  T.  To  prove  that  bC 
is  perpendicular  to  tt,  the  //-trace  of  T. 

2 


18 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Let  fall  a  perpendicular,  Bb,  from  B  to  H.  The  lines  BC  and 
Bb  determine  a  plane  perpendicular  to  both  H  and  T,  and 
therefore  to  their  intersection,  tt.  As  the  plane  is  perpendicular 
to  tt,  its  intersection  bC  will  be;  that  is,  the  ^"-projection  is 
perpendicular  to  the  //-trace.  The  same  could  be  proved  for 
the  F-projection  and  F-trace. 


FIG.  12. 

Theorem  XXV. — Conversely,  a  line,  whose  projections  are 
perpendicular  to  the  respective  traces  of  a  plane,  is  perpendicular 
to  the  plane,  with  one  exception. 

Exception. — If  the  plane  is  parallel  to  GL,  the  line,  whose 
projections  are  perpendicular  to  its  traces,  may  not  be  perpen- 
dicular to  the  plane.  Prove  this  by  the  P-trace  of  such  a  plane, 
and  the  P-projection  of  such  a  line,  rotated. 

26.  Theorem  XXVI. — A  line  in  any  plane  pierces  H  in  the 
H-trace  of  the  plane,  unless  the  line  is  parallel  to  H.     Similarly, 
it  will  pierce  V  in  the  F-trace  of  the  plane  in  which  it  lies. 
Prove. 

Theorem  XXVII. — If  a  line  is  parallel  to  H,  it  is  parallel  to  the 
H-trace  of  any  plane  in  which  it  lies.  Similarly  for  F.  Prove. 

Theorem  XXVIII. — Any  plane  that  is  passed  through  one  of 
two  parallel  lines,  either  contains  or  is  parallel  to  the  other. 

Let  the  student  prove  this  with  a  figure. 

EXERCISES  IN  REVIEW 

27.  Directions. — All    exercises    will    be    drawn     in    a    space 
6  in.  X  7  in.,  with  GL  horizontally  across  the  middle. 


DEFINITIONS,  NOTATION,  PRELIMINARY  THEOREMS         19 

Letter  all  points  carefully,  using  small  letters  for  all  projections, 
a',  b',  etc.,  for  F-projections,  a,  b,  etc.,  for  /^-projections,  ap,  bp, 
for  P-projections.  ^ 

Use  the  following  conventions  for  all  lines,  etc. 

Given  lines  —  solid,  medium,  j  _  fa" 

Required  lines  —  solid,  heavy  —  «—_  «—  ^—   s1./' 

Auxiliary  lines  —  solid,  light  _  very  fine. 

Invisible  lines  —  dotted,  medium  —  fa" 

Projectors  —  dotted,  light  —   very  fine. 

Paths  of  Revolution  —  the  same. 

Traces  of  Given  planes  —  medium  _  _ 

Traces  of  Required  planes  —  heavy  ^_^  _ 

Traces  of  Auxiliary  planes  —  light  _  very  fine. 

All  planes  are  considered  transparent,  and  there  will  be 
no  invisible  lines  until  problems  involving  solids  are  given. 
1  1  and  J_  are  abbreviations  for  parallel  and  perpendicular. 


a!  w/ 


FIG.  13. 

28.  Through  A(l  +  !$-!)  draw  a  line  AB  ||  C(2  +  $+l)  D(3|-l-i). 

29.  Through  M(4|-l+2)  draw  a  line  MN  ||  and  equal  to  K(l 


30.  From  D(5  +  2-l)    draw  two  lines  intersecting  the  line  A(2  — 
B(3i~  \  —  $)  in  points  other  than  A  and  B. 

31.  Draw  two  lines  AB  and  CD  intersecting  in  the  point  E(3  —  2  —  $). 
Connect  these  two  lines  with  a  third. 

32".  Draw  four  lines  intersecting  A  (2  —  1  +  f  )  B(3  —  2  +  f  )  at  right  angles  to  it. 

33.  Through  O(4  +  2-l)  draw  a  line  _L  to  T(l  +2)4(1-3). 

34.  Draw  the  traces  of  a  plane  S  J_  to  A(3+2-li)B(4  +  l  -$). 

35.  Is  M(3  +  l|-i)N(3  +  $-l|)   ±  to  T(  +  l)oo(-|)?     If  not,  draw  the 
traces  of  a  plane,  S,  that  is  J_  to  MN. 


20  PRACTICAL  DESCRIPTIVE  GEOMETRY 

36.  Draw     the     P-projections     of     the    following    points:    A(£  +  l  +  l), 
B(li-2  +  i),    C(2i  +  H-f),    D(3-l-i),    and  E(3|,  0  +  i).     Place 
P  at  4$. 

37.  Find  the  P-projections  of  the  following  lines:  M(f  +  1J  +  l)N(J  — J  — f) 
andO(2  +  H-i)P(2-l-i).     Profile  at  4^. 

38.  Find    the    P-projections    of    the    lines    E(l-l  — 1)F(1  — J  +  l)  and 
G(2  +  2  +  f)H(2  +  l-l),  Pat4i 

39.  Write  the  notation  of  the  planes  in  Fig.  13,  and  describe  them.     Use 
your  own  dimensions. 

40.  Draw  through  O(2  +  l~i)  a  line,  OP,  2"  long,  \\toH,  and  oblique  to 
V,  and  a  line  ON,  2£"  long,  1 1  to  V,  and  oblique  to  H. 

41.  Let  A(2,  0-l)B(2i,  0-2)  be  the  diagonal  of  a  cube  resting  on  H. 
Draw  the  projections  of  the  cube,  and  the  traces  of  the  six  planes 
bounding  it. 


CHAPTER  II 
PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES 

28.  General   Hints. — In   the   solution   of   the   problems   that 
follow,  we  (1)  analyze  the  operations  in  space,  or  on  paper,  as  the 
case  may  require,  (2)  offer  proof  of  the  correctness  of  this  analy- 
sis, when  it  seems  necessary,  and  (3)  show  the  construction  of  a 
typical  problem  "on  the  flat."     The  analysis  is  the  important 
part  to  learn,  and  should  be  thoroughly  learned,  not  through 
memorizing  the  words  or  the  figure,  but  by  visualizing  the  con- 
dition in  space,  and  imagining  the  process.     The  analysis  fur- 
nishes the  "machinery"  for  the  work  on  the  flat,  and  the  worker 
should  be  adept  with  his  tools.     This  can  be  done  only  by 
working  exercises,  of  which  there  is  an  ample  supply  nri— 
No  attempt  should  be  made  by'' 

exercise  in  space,  but  the  work  < 
and  worked  according  to  the  anab    ; 
will  be  clear. 

29.  Problem  i. — To  find  the  points 
andV. 

Analysis. —  (1)     Extend  the  F-pi'Oj^iI^   to    GL. 
this  point  erect  a  perpendicular  (projector)    to  the  //"-projection, 
or   its  -extension.     (3)    The    point    of   this    intersection  is  the 
//"-piercing  point. 

Interchanging  H  and  V  in  the  analysis  will  obtain  the 
F-piercing  point. 

Proof. — Since  all  points  in  H  have  their  F-projection  in  GL, 
this  point  must  be  in  H,  and  therefore  the  point  where  the  line 
pierces  H. 

Construction. — Let  AB  (Fig.  14)  be  any  oblique  line. 

1.  Extend  a'b'  to  c'  on  GL. 

2.  Draw  the  projector  to  c  on  the  extension  of  ab.     G  is  the 
//-piercing  point. 

3.  Extend  ab  to  d  on  GL. 

4.  Draw  the  projector  to  d'  on  the  extension  of  a'b'.     D  is 
the  F-piercing  point. 

21 


22 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Axioms. —  1.    A  line  parallel  to  V  has  no  F-piercing  point, 
therefore  its  //-projection  must  be  parallel  to  GL. 

2.     A  line  parallel  to  H  has  no  //-piercing  point,  therefore 
its  F-projection  must  be  parallel  to  GL. 


FIG.  14. 


30.  Problem  2. — To  find  the  H-  and  V -piercing  points  of  a  line 
in  a  profile  plane. 

Analysis. —  1.  Project  the  line  on  P,  located  (preferably)  on 
the  right.  See  Fig.  8a. 

2.  Revolve  P  about  its   F-trace  into    F.     (Determine  the 
P-projections  of  two  points  of  the  line.     See  Fig.  8a.) 

3.  Extend  the  P-projection  of  the  line  to  H  and  F,  which 
are  represented  respectively  by  GL  and  the  P-trace. 

4.  Project  these  points  back  to  the  original  projection  of  the 
line,  or  its  extension. 

Proof. —In  rotating  P  into  F,  all  points  in  H  will  project  on 
GL,  and  all  points  in  F  will  project  on  the  F-trace  of  P. 

Construction. — Let  A B  (Fig.  15)  be  a  line  in  a  profile  plane. 

1.  Project  it  on  P(apbp),  and  revolve  it  as  shown. 

2.  Extend   apbp  .to   H(GL),    and    V(Pp'),   to   the  points   cp 
and  dp. 

3.  Revolve  Cp  down  to  Pp  and    project  it  back  to  ab  (ex- 
tended), at  C,  the  //-piercing  point. 

4.  Project    dp   back   to    d'    on    a'b'    (extended),  giving  the 
F-piercing  point. 

Fig.  15a  shows  the  problem  when  A  is  in  //  and  B  in  ///, 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       23 

Rules. — 1.  The  77-proj  actions  that  lie  below  GL  (points  in 
/  and  7V)  revolve  counter-clockwise  and  return  clockwise. 

2.  The  ^-projections  that  lie  above  GL  (points  in  77  and  777) 
revolve  clockwise  and  return  counter-clockwise. 


FIG.   15. 


FIG.   15a. 


3.  All  F-projections  move  parallel  to  GL. 

Axiom. — Lines  in  profile  planes,  when  projected  on  P,  revolved 
into  V,  are  shown  in  their  true  length,  and  in  their  true  angles 
to  H  and  V. 


24 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


EXERCISES 

31.  Find  the  points  in  which  H  and  V  are  pierced  by  the  following  lines, 
or  their  extensions.     Tell  what  angles  they  traverse. 
42. 
43. 
44. 
45. 
46.  E(l,  0  +  li)F(2i,  0- 

Note  . — For  the  following  exercises  one  P-plane  at  4^"  from  the  margin 
will  suffice  for  each  exercise. 

47. 
48. 
49. 
50. 
51.  P(i+ii  +  i)D(l+i+l)j  E(2-i-l)F(2-2-±). 


FIG.  16. 

32.  Problem  3. — To  find  the  piercing  point  of  a  line  with  a 
profile  plane,  and  show  it  in  P-projection. 

Analysis. — 1.  The  point  in  which  the  line  pierces  P  will  have 
its  H-  and  F-projections  where  the  P-traces  cut  the  H-  and 
F-projections  of  the  line. 

2.  Revolve  this  point  as  in  Problem  2. 

Proof. — P  is  a  projecting  plane  on  H  and  V,  hence  any  point  in 
the  plane  will  be  projected  on  its  traces.  (Theorem  X,  Art.  18.) 

Construction. — Let  it  be  required  to  find  the  P-piercing  point 
of  AB  (Fig.  16). 

1.  Extend  a'b'  to  the  P-trace  at  c'. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       25 

2.  Extend  ab  to  the  P-trace  at  c.     c  is  the  required  point. 

3.  Revolve  P  into  V,  locating  cp. 

33.  Problem  4. — To  assume  a  line  in  a  plane. 
Analysis. — 1.  Assume  a  point  in  the  F-trace. 

2.  Assume  a  point  in  the  H -trace. 

3.  Connect  the  respective  projections  of  the  points. 

4.  The  line  will  be  contained  in  the  plane. 

Proof. — If  any  two  points  lie  in  a  plane,  the  line  joining  them 
will  lie  wholly  in  the  plane. 

Construction. — Let  t'Tt  (Fig.  17)  be  any  plane. 

1.  Assume  a'  on  tT,  a  will  be  in  GL.     (Theorem  IX,  Art.  18.) 

-2.  Assume  b  on  tT,  b'  will  be  in  GL. 

3.  Draw  a'b'  and  ab. 

4.  AB  will  be  contained  in  T. 


FIG.  17. 

34.  Second  Case. — If  it  be  desired  to  draw  a  line  in  T  parallel 
to  H,  assume  a  point  c'  on  t'T,  and  draw  c'd'  parallel  to  GL. 
Project  c'to  c  on  GL,  and  draw  cd  parallel  to  Tt.     (Theorem  V, 
Art.  15,  and  Theorem  XXVII,  Art:  26.) 

35.  Problem  5. — One  projection  of  a  point  in  an  oblique  plane 
being  given,  to  find  the  other  projection. 

Analysis. — 1.  Through    the    given    projection    draw    a    line, 
assuming  it  to  be  one  projection  of  a  line  in  the  given  plane. 

2.  Locate  the  other  projection  of  the  line.     (Problem  4.) 

3.  Project   the   given  point   to   the   other   projection   of   the 
assumed  line,  and  this  will  be  the  required  point. 

Proof. — If  a  line  lies  in  a  plane,  all  its  points  are  in  the  plane. 
Construction. — See  Fig.  17. 

1.  Assume  m'  or  n'. 

2.  Through  m'  draw  a'b',  or  through  n'  draw  c' 


26 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


3.  Locate  -ab  or  cd  (Problem  4). 

4.  Project  m'  to  m  on  ab,  or  n'  to  n  on  cd.     M  and  N  will 
lie  in  T. 

36.  Hint. — -It  is  usually  more  satisfactory  and  speedy  to  use 
a  parallel  line,  like  CD,  rather  than  an  oblique  line,  like  AB. 
Sometimes,  as  in  the  case  of  a  plane  parallel  to  GL,  the  oblique 
line  only  can  be  used  conveniently. 

Note. — If  a  point  is  to  be  assumed  in  a  projecting  plane,  the  point  may 
be  indeterminate;  that  is,  if  the  assumed  projection  is  taken  on  the  pro- 
jecting trace,  there  may  be  an  infinite  number  of  points  having  that  pro- 
jection, and  lying  in  the  plane. 

37.  Problem  6.— To  find  the  traces  of  a  plane  determined  by 
three  given  points. 

Analysis. — 1.  Join  the  points,  two  and  two,  by  lines. 

2.  Find  the  piercing  points  of  the  lines. 

3.  Draw  the  //-trace  through  the  //-piercing  points,  and  the 
F-trace  through  the  F-piercing  points. 


't 


FIG.  18. 


Proof. — The  lines  huve  their  piercing  points  in  the  respective 
traces  of  the  plane,  therefore  the  lines  (and  all  their  points)  lie 
in  the  plane  thus  determined. 

Construction. — Fig.  18  shows  three  points,  A,  B  and  C. 

1.  Join  the  respective  projections  of  AB  and  BC. 

2.  Find  the  piercing  points,  D,  E,  F  and  G. 

3.  Through  d'  and  e'  draw  t't',  and  through  f  and  g  draw  tt. 
These  will  be  the  required  traces. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       27 

Check. — 1.  The  line  BC  should  have  its  piercing  points  in  the 
respective  traces  of  T,  and 

2.  the  traces  should  meet  in  a  point  on  GL,  or  be  parallel  to  GL. 

Hints. — 1.  Sometimes  the  points  may  be  so  placed  that  the 
piercing  points  are  not  all  to  be  found  within  the  limits  of  the 
problem.  In  such  a  case,  an  auxiliary  line  may  be  drawn  from 
any  of  the  three  points  to  any  point  on  the  line  joining  the  other 
two.  For  instance,  if  such  a  proceeding  were  necessary,  a  line 
could  be  drawn  from  B  to  any  point,  M,  on  the  line  AC.  BM 
would  lie  in  the  required  plane,  and  its  piercing  points  would 
help  to  determine  the  traces. 

2.  Ordinarily,  only  three  piercing  points  are  necessary  to  be 
found,  because  when  the  #-trace  is.  found,  its  intersection  with 
GL  may  be  joined  to  one  of  the  F-piercing  points. 

38.  If  two  intersecting  lines  are  given. — This  makes  no  change 
in  the  problem. 

39.  If  two  parallel  lines  are  given.— Find  their  piercing  points 
and  draw  the  traces,  as  in  the  previous  case. 

40.  If  a  point  and  a  line  are  given. — The  point  may  be  con- 
nected with  any  point  in  the  line,  thus  making  two  intersecting 
lines,  or  through  the  point  a  line  may  be  .drawn  parallel  to  the 
given  line,  thus  making  two  parallel  lines.     (Theorem  XIX, 
Art.  24.) 

EXERCISES 

41.  Draw  the  H-  ,  V-  and  P-projections  of  the  following  lines.     Place  P 
at  4£"  in  each  exercise. 

51. 
.52. 

53. 

54. 

.Locate  the  following  lines  in  the  planes  given: 

55.  AB,  any  oblique  line  in  T  (4£+ 2)2(4$ -1$). 

56.  C(2  +  2,  x)D(3  +  l,  y)in  T  (Ex.  55).     State  values  of  x  and  y. 

57.  E(2-l,  x)F(l-i,  y)  in  T  (Ex.  55).     State  values  of  x  and  y. 

58.  GH,  a  line  ||  to  H,  1±"  above  H  in  T  (Ex.  55). 

59.  KL,  a  line  ||  to  V,  I"  behind  V  in  T  (Ex.  55). 

60.  MN,  a  line  in  ///,  ||  to  t'T  (Ex.  55). 

61.  0(4  +  1,  x)P(2,  0,  0)  in  T  (Ex.  55).     State  value  of  x. 

62.  Find  the  locus  of  all  points  in  T  (Ex.  55),'  I"  above  H,  also  of  those 
\"  below  //. 

63.  Find  the  locus  af  all  points  in  T  (Ex.  55),  I"  behind  V,  also  of  those 
1"  in  front  of  7. 


28  PRACTICAL  DESCRIPTIVE  GEOMETRY 

64.  Locate  the  following  lines  in  S(  +  2)oo(  —  1$):  A(l  +  1$,  x)B(2  +  J,  y), 
C(2i,  z  +  i)D(3$  +  l,  q),  and  EF,  the  locus  of  all  points  $"  in  front  of 
V.     (Hint.     It  is  not  necessary  to  use  a  profile  plane  to  obtain  any 
of  these  lines,  though  it  may  be  found  useful  for  the  line  EF.) 

65.  Locate  the  following  lines  in   R(  +  l)oo(-f-2):  G(1  +  J,  x)H(2,  y-i), 
K(3,  z  +  l)L(4  +  l$,  q)  and  MN,  locus  of  all  points  \"  below  H. 

Locate  the  following  points  in  the  planes  given: 

66.  A(4  +  l,  x),    B(3  +  2,   y),   C(2,   z-2),   and   D(l-2,   q)   in   the  plane 


67.  E(2,  x  +  i),  -F(3i,  y-1),   G(4,  z  +  i),  and  H(5,  q-|)   in  the  plane 
R(l+2)3i(6-2). 

68.  M(l,  x  +  i),  N(2-l,  y),  and  O(4,  z-1)  in  the  plane  S(  +  l|)oo(-l). 
Find  the  traces  of  the  planes  determined  by  the  following  points  and 

lines: 

69.  A(3  +  l-i),  B(3i  +  i  +  H),  andC(4- 

70.  D(2—  2  +  f),  E(3i  +  i-lJ),  an 

71.  K(2  +  li-i),  L(3  +  i-|),  and 

72.  N(2  +  l-l),  O(3-l 

73.  A(li  +  3  +  l)B(2i  +  l-l),  and 

74.  The  triangle  D(2  +  l 

75.  The    parallelogram    K  (14  +  1  -f)  L  (2  +  If  -f)  O  (2|  +|-lf) 

1-lf). 

76.  With  the  line  A  (2,  0-1)B(2|,  0-2)  as  a  base  diagonal,  erect  a  cube 
standing  on  H.     Find  the  traces  of  all  the  planes  bounding  the  cube. 

77.  With  the  line  C(3,   0-2)D(34,   0-1)   as  a  base  diagonal,   erect  a 
square  pyramid  of  2"  altitude.     Draw  its  projections  and  the  traces 
of  the  four  planes  of  the  slanting  facse. 

42.  Problem  7.  —  To  pass  a  plane  through  a  given  point  par- 
allel to  two  given  lines. 

Analysis.  —  1.  Draw  through  the  given  point  two  lines,  parallel 
respectively  to  the  two  given  lines.  (Theorem  XIX,  Art.  24.) 

2.  Pass  a  plane  through  the  lines  thus  drawn. 

Proof.—  Referring  to  Theorem  XXVIII,  Art.  26,  we  see  that 
the  plane  thus  found  must  be  parallel  to  both  of  the  given  lines, 
because  it  contains  lines  parallel  to  each  of  them. 

Construction.  —  Let  A,  Fig.  19,  be  the  given  point,  and  BC 
and  DE  be  the  given  lines. 

1.  Through  ar  draw  mV  parallel  to  bV,  and  o'p'  parallel 
to  d'e'.     Through  a  draw  mn  parallel  to  be,  and  op  parallel 
to  de. 

2.  Find  the  H-  and  F-piercing  points  of  MN  and  OP. 

3.  Through  these  piercing  points  draw  the  traces  of  T. 

Note.  —  If  the  two  given  lines  are  parallels,  an  infinite  number  of  planes 
may  be  passed  through  the  point  parallel  to  the  two  lines,  and  the  problem 
is  therefore  not  definite. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       29 


43.  Problem  8.  —  Through  a  given  line  to  pass  a  plane  parallel 
to  another  given  line. 

Analysis.  —  1.  Through  any  point  in  the  given  line  draw  a  line 
parallel  to  the  second  given  line. 

2.  Pass  the  required  plane  through  the  two  intersecting  lines 
thus  drawn. 

Let  the  student  supply  the  proof  and  make  the  construction. 

Note.  —  If  a  plane  be  passed  through  the  second  line  parallel  to  the  first, 
this  plane  will  be  parallel  to  the  first  plane. 


EXERCISES 

4.  Pass  planes  through  the  following  points  parallel  to  the  lines  given. 
'  78.  Through   O(3  +  l-l)  ||  to  A(l  +  l$-l)B(2  +  i-2)   and  C(4  +  l-l) 


79.  Through  M(4-i  +  l)  ||  to  E(1J  +  H-li)F(2-i  +|,  0),  and 
K(4  +  l*  +  2). 

80.  Through  A(3i  +  2-l)B(4  +  l  -£)  pass  the  plane  T  ||  to 
D(2  +  li-i). 

81.  Through   E(3  +  1£-£)F(4  +  1  -f)    pass   a   plane  S   ||   to 
H(li+H-i). 

82.  Through    K(3$  +  l-|)L(5+2-l)    pass   a    plane    ||    to    M(3-l  +  l) 


45.  Problem  9.  —  To  revolve  a  point  in  space  into  H  about  a 
line  contained  in  H. 

Definition  of  Revolution.  —  To  revolve  a  point  about  a  line 
(called  its  axis)  means  to  move  the  point  in  a  plane  perpen- 
dicular to  the  axis,  keeping-  it  at  a  constant  distance  from  the 
axis.  The  path  of  the  point  is  a  circle,  and  the  center  of  the 
circle  is  the  point  where  the  axis  pierces  the  plane  of  revolution. 

Process.  —  Examine  Fig.  20,  which  is  a  picture  of  the  two 
planes  of  projection,  H  containing  the  line  BC,  and  the  point 


30 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


A  being  located  in  space.  The  projections  of  A  are  shown  on 
V  and  H.  The  circle,  in  which  A  is  to  revolve,  lies  in  a  plane 
perpendicular  to  BC  (by  hypothesis,  see  foregoing  definition), 
and  is  therefore  perpendicular  to  H.  (Prove  by  Solid  Geom- 
etry.) This  plane  is  therefore  a  projecting  plane,  and  contains 
not  only  A,  but  its  //"-projection,  a;  and  its  //-trace  runs  through 
the  projection,  a,  perpendicular  to  BC,  and  crosses  BC  at  x, 
the  center  of  the  circle  of  revolution.  Ax  is  then  the  radius  of 
revolution.  Ax  is  the  hypothenuse  of  the  right  triangle  Axa. 
A  will  revolve  into  H  at  the  point  at,  which  is  upon  the  line  ax 
a  distance  equal  to  the  hypothenuse  of  Axa.  The  side  Aa  of 
Axais  equal  to  a'y,  by  Theorem  I,  Art.  9.. 


FIG.  20. — The  revolution  of  a  point. 

Analysis. — 1.  Draw  the  trace  of  the  plane  of  revolution 
through  the  //-projection  of  the  point,  and  perpendicular  to  the 
axis. 

2.  Construct  the  right  triangle,  as  follows:  First,  its  base  is 
the  distance  of  the  //-projection  of  the  point  to  the  axis,  and 
second,  its  altitude  the  distance  of  its  F-projection  from  GL. 

3.  Lay  off  the  length   of  the  hypothenuse   of  this   triangle 
along  the  trace  of  the  plane  of  revolution,  measuring  from  the 
axis  in  either  direction. 

4.  This  point  will  be  the  revolved  position  of  the  point. 
Proof.— The  process,  already  given,  contains  the  proof. 
Construction. — Refer  to  Fig.  21. 

Let  A  be  the  point,  and  BC  the  given  line  in  H. 

1.  Through  a  draw  a  line  perpendicular  to  be. 

2.  Construct  a  right  triangle  axz  using  ax  and  a'y  as  sides. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       31 

3.  Lay  off  the  hypothenuse,  xz,  along  xa  (extended  in  either 
direction)  from  the  point  x. 

4.  aA  will  be  the  required  position. 

Notes. — 1.  To  revolve  a  point  into  V  about  a  line  in  V,  make  an  inter- 
change of  H  and  V  in  the  foregoing. 

2.  If  the  point  is  directly  over  the  axis,  the  base  of  the  triangle  becomes 
zero,  and  the  altitude  and  hypothenuse  w'll  be  equal. 

3.  The  "  trace  of  the  plane  of  revolution"  and  "  the  projection  of  the  path 
of  revolution"  will  be  abbreviated  hereafter  to  "the  path  of  revolution." 


46. 


82 


EXERCISES 

and    B(5-1 


into   H    about  C(l,  0-2) 


Revolve 
D(5,  0  +  2). 

83.  Revolve  E(3  +  lf  +  l)    and  F(4-i-l$)    into    V   about   G(l-2,    0) 
H(5  +  2,  0). 

84.  (a)  Revolve  K(l+2-i),  L(2  +  *•+!£),  M(3-l-l),  and  N(4-*  +  2) 
into  H  about  GL. 

(b)  Revolve  them  into  V  about  GL. 

85.  (a)  Revolve  M(2  +  1-1)  into  V  about  O(l  +  l,  0)N(3  +  2,  0). 
(b)  Revolve  A  (4  +  2  +  )  into  H  about  B(3$,  0-2)0(5,  0-1). 

86.  Locate  D(3  +  l,  x)  and  E(4,   y-f)  in  1(5  +  3)1(5-2).     Revolve  D 
and  E(l)  about  t'T  into  V,  (2)  about  Tt  into  H. 

87.  Locate  F(2  +  2,  x)  and  K(3£,  y-lf)  in  S(  +  l£)oo(-l).     Revolve  F 
and  K  about  the  //-trace  of  S  into  H. 

88.  Locate  M(2  +  l,  x)  and  N(4,   y  —  $)   in  S(  +  li)oo(  +  £),   and  revolve 
them  into  V  about  the  V-trace  of  S. 

89.  Revolve  two  points,  O  and  P,  both  in  the  F-trace  of  T  (Ex.  86).  O 
being  \"  and  P  being  \\"  above  H,  about  the  //-trace  of  T  in 

90.  Revolve  A(2  +  li,  0)   about  the  //-trace  of  S(  +  li)co(-f)    in1 
Revolve  B(4,  0-f)  about  the  F-trace  into  V. 


32 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


P. 


91.  Revolve  the  triangle  D(2  +  l- i)E(3-i-2)F(3i  +  i-l)   about  the 
//-trace  of  its  plane  into  H.     Will  this  be  its  true  size? 

92.  Revolve     the     parallelogram     K(2  +  l-lJ)L(3+lf-l)O(4+ 1,  0) 
N(3  +  ^  —  |)  about  its  F-trace  into  F.     Will  this  give  its  true  size? 

47.  Problem  10. — To  revolve  a  point  through  any  angle  about 
a  line  perpendicular  to  H  or  V. 

Analysis. — (About  a  line  perpendicular  to  //.) 

1.  The  point  moves  in  a  circle  parallel  to  H,  which  is  projected 
on  H  in  a  circle  of  the  same  size. 

2.  Draw  the  circle,  using  as  radius  the  distance  from  the  H- 
projection  of  the  point  to  the  //-projection  of  the  line. 

3.  Draw  the  F-projection  of  the  circle.     This  will  be  a  line 

through    the    F-projection    of    the    given 
point,  and  parallel  to  GL.     Why? 

4.  Lay  off  the  required  arc  on  the  circle 
from  the   original   position    of    the    point 
(//"-projection),  locating   the  //-projection 
of  the  revolved  position  of  the  point. 

5.  Project  this  point  to  the  F-projection 
of  the  circle. 

Proof. — The  proof  of  this  depends  upon 
the  truth,  that  the  projections  of  the  circle 
are  as  given  in  the  analysis.  Make  a  draw- 
ing and  give  this  proof. 

Construction.— Let  0  (Fig.  22)  be  the 
given  point,  AB  the  given  axis,  perpen- 
dicular to  //,  and  6  the  given  angle. 

1.  Draw  the  //-projection  of  the  circle 
with  a  as  its  center  and  ao  its  radius. 

2.  Draw  the  F-projection  of  the  circle,  through  o' ,  parallel 
to  GL. 

3.  Move  o  through  the  given  angle,  0,  to  or 

4.  Project  ot  to  o\,  giving  both  projections  of  the  point  in  the 
required  revolved  position. 

48.  Problem  11. — To  revolve  a  line  through  any  given  angle 
about  a  line  perpendicular  to  H  or  V. 

Analysis. — 1.  As  the  line  is  the  sum  of  all  points  in  it,  take 
any  two  points  of  it  and  revolve  them  about  the  axis,  as  in 
Problem  10. 

2.  Join  these  revolved  points,  and  the  resulting  projections 
will  be  projections  of  the  line. 


/ 


FIG.  22. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     33 

49.  EXERCISES 

93.  Revolve  A(li  +  l  —  2)  90°  counter-clockwise  about  M(l,  0-1)  N(l  + 
2-1),  and  B(3$  +  2-li)  90°  clockwise  aboutP  (4  +  1,  0)  O(4  +  l-2). 
Also  revolve  B  until  it  comes  to  H  . 

94.  Revolve  C(2  +  l|  +  l)   45°  clockwise    about    K(l  +  1-2)  L(3  +  l  -2), 
and  M(4i  +  $-2)  N(5£  +  l|-l)   135°  counter-clockwise  about  D(4, 
0-1)  E(4  +  2-l). 

95.  Revolve  A(2  +  i-l)  B(4  +  l£-2)  about  P(3  +  l-0)  O(3  +  l-3)  into 
a  position  parallel  to  H. 

96.  Revolve  E(2  +  2-l)F(3  +  £-2)   75°    clockwise    about    K(l  +  l-l|) 


50.  Problem  12.  —  To  find  the  true  length  of  a  line  in  space. 

Discussion.—  A  line  oblique  to  a  plane  is  not  projected  in  its 
true  length  on  that  plane.  To  be  projected  in  its  true  length, 
a  line  must  be  parallel  to  that  plane,  or  be  contained  in  it.  See 
Theorems  V-X. 

It  follows,  then,  that  any  oblique  line,  in  order  to  be  measured, 
must  be  moved,  without  changing  its  actual  length,  into  a 
position  parallel  to  or  in  H  or  V. 

Case  I.  —  By  revolving  parallel  to  V. 

Analysis.  —  1.  Through  any  point  in  the  given  line  (preferably 
one  of  its  extremities)  draw  an  axis  perpendicular  to  H. 

2.  Revolve  the  line  about  this  axis  (Problem  11)   until  the 
//-projection  is  parallel  to  GL. 

3.  The  line  will  then  be  parallel  to  V,  and  its  7-projection 
will  be  the  true  length  of  the  line. 

Proof.  —  The  line  remains  unchanged  in  revolving  about  its 
axis,  because  every  point  revolves  about  the  axis  in  a  series  of 
parallel  circles  through  the  same  angle.  This  means  that  every 
point  remains  in  the  same  relation  to  the  axis  throughout  the 
revolution,  and  therefore  all  points  in  the  line  remain  in  the 
same  relation  to  each  other,  and  the  line  is  unchanged. 

(Note.  —  The  length  of  the  //-projection  is  not  changed,  but  that  of  the 
y-projection  is  changed.) 

Construction.  —  Let  AB  (Fig.  23)  be  the  given  line. 

1.  Let  CA  be  the  axis,  perpendicular  to  H. 

2.  Revolve  b  in  a  circle  about  a  as  a  center,  until  abx  is 
parallel  to  GL. 

3.  Move  b'  parallel  to  GL,    until  it  intersects    the  projector 
from  br 

4.  Draw  a'b'j,  which  is  the  true  length  of  AB. 

3 


34 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Check. — Revolve  AB  about  an  axis  perpendicular  to  V, 
until  it  is  parallel  to  H ,  and  this  result  should  be  equal  to  the 
first  result. 

51. — Case  II. — By  revolving  into  H. 

Analysis. — 1.  Let  the  axis  be  the  //-projection  of  the  given 
line. 

2.  Revolve  the  line  about  this  axis  into  the  plane  of  the  axis, 
H.  (Problem  9,  Note  2.) 

The  line  will  be  in  H,  and  therefore  projected  in  its  true 
length  on  H.. 

Proof. — By  using  the  H  -projection  of  the  line  for  the  axis, 
we  make  sure  that  the  line,  revolving  about  the  axis,  can  be 
brought  into  H,  in  which  the  axis  lies. 


FIG.  23. 


FIG.  24. 


The  analysis  might  be  stated  thus:  Fold  the  projecting  plane 
of  the  given  line  into  H,  about  its  line  of  intersection  with  H. 

Construction. — Let  AB  (Fig.  24)  be  the  given  line,  and  ab 
the  axis. 

1.  Lay  off  perpendiculars  to  ab  from  a  and  b.     (See  Problem 
9,  Note  2.) 

2.  Measure  off  on  these  perpendiculars  a'x  and  b'y,  giving  the 
points  at  and  br 

3.  Connect  at  and  br     This  will  be  the  required  length. 

Notes. — 1.  The  line  pierces  H  in  its  77-projection,  and  therefore  the  //-pierc- 
ing point  of  the  line  lies  in  the  axis,  and  will  remain  stationary  in  revolution. 
This  fact  may  be  used  as  a  check  on  the  work. 

2.  If  the  ends  of  the  line  are  on  opposite  sides  of  the  plane  into  which 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       35 

they  are  revolved,  the  perpendiculars  will  have  to  be  laid  off  on  opposite 
sides  of  the  axis. 

Referring  to  Fig.  24a,  we  have  AB,  a  line  running  through  V  from 
point  in  /  to  a  point  in  II.  The  F-piercing  point  is  at  o  between  A  and  B. 
Taking  a  portion,  BC,  of  AB,  lying  entirely  on  one  side  of  V,  and  revolving 
it  to  b/1c/l,  we  find  that  \)\,  c'v  and  o'  lie  in  one  line,  and  continuing  that 
line,  we  find  that  it  meets  A  in  revolved  position  at  a'p  on  the  opposite  side 
of  the  axis. 

3.  In  both  of  the  foregoing  cases  of  Problem  12  the  planes  V  and  H  are 
interchangeable. 


FIG.  24a. 

52.  Case  III. — When  the  line  is  in  a  profile  plane. 
Analysis. — 1.  Project  the  line  on  P  and  revolve  into  V,  as 

described  in  Problem  2. 

2.  The  P-projection;  thus  revolved,  is  the  true  length  of  the 
line. 

Proof.— The  line  is  parallel  to  P,  and  is  therefore  projected  on 
P  in  its  true  length. 

Construction. — Exactly  the  same  as  employed  in  Problem  2, 
Figs.  15  and  15a. 

Note. — Points  in  II  and  ///  will  be  rotated  to  the  right  side  of  P. 

53.  Problem  13. — To  lay  off  a  given  distance  on  a  given  oblique 
line. 

Analysis. — 1.  Revolve  any  definite  portion  of  the  line  to  show 
its  true  length,  as  in  Problem  12. 

2.  Lay  off  the  given  distance. 

3.  Counter  revolve  the  limiting  points  to  the  original  position 
of  the  line. 

Construction. — Use  Fig.  24.     Let  it  be  required  to  lay  off  an 
inch  from  A  on  AB. 
1.  Revolve  AB  to  axbr 
2    Let  a,^  be  1  in.     Lay  it  off  from  a^ 


36  PRACTICAL  DESCRIPTIVE  GEOMETRY 

3.  ct  being  the  other  extremity  of  the  line,   must  now  be 
revolved  back  with  the  line  into  the  original  position.     As  all 
points  of  the  line  revolve  in  parallel  circles,  the  return  revolution 
will  be  made  exactly  as  the  outward  revolution.     The  paths  of 
revolution  are  perpendicular  to  ab,  therefore  C  will  move  back 
to  c'c,  the  same  as  A  and  B  moved  into  H,  in  a  line  whose 
projection  is  perpendicular  to  the  axis. 

4.  AC  is  then  a  part  of  AB  1  in.  long. 

Note.  —  Any  method  of  measuring  a  line  may  be  used  in  laying  off  this 
distance. 

Axiom.  —  If  a  line  be  divided  into  any  number  of  segments, 
the  projections  of  these  segments  will  be  proportional  to  the 
segments  in  space.  Thus,  to  bisect  a  line,  bisect  its  projections, 
etc.  Prove  this  by  Plane  Geometry. 

54.  EXERCISES 

97.  Measure  A(l$  +  $-$)  3(4  +  2-1$),  (1)  by  revolving  ||  to  V;  (2)  by 
revolving  into  H. 

98.  Measure  C(2  +  $-2)   D(4  +  2-£),    (1)  by  revolving  ||   to  H;  (2)  by 
revolving  into  V. 

99.  Measure  E(2  +  $  +  $)  F(4  +  2  +  2),   (1)  by  revolving  ||   to   V;  (2)  by 
revolving  into  H  . 

100.  Measure  G(l+2  +  l)  H(4-l-l$),  (1)  by  revolving  ||  to  H;  (2)  by 
revolving  into  V. 

101.  Measure  K(l  +  !$-!)  L(4-l+2$),  (1)  by  revolving  ||  to  V;  (2)  by 
revolving  into  H  . 

102.  Measure  M(2-$-$)  N(4-2$-2$),  (1)  by  revolving  ||  to  H;  (2)  by 
revolving  into  V. 

Find  the  true  length  of  the  following  lines.     Place  the  Profile  plane  at  4$". 

103.  A(l  +  2-l)  B(l+$-l),  and  0(2  +  1  +  1$)  D(2-l  +  $). 

104.  E(l+l  +  i)  F(l  +  l$-$),  and  G(2  +  l+$)  H(2-i-2). 

105.  K(l  +  l-l)  L(l-$  +  l),  and  M(2-$  +  f)  N(2-2+±). 

106.  O(l-$-l)  P(l+2-$),  and  A(2-l$  +  i)  B(2-$-l$). 

107.  C(l  +  l$  +  i)  D(l+f  +  1),  and  E(2-$-l)  F(2-2-$). 

108.  Lay  off  on  the  line  A(2  +  2-l$)  B(4  +  $-$)  the  point  C,  1$"  distant 
from  A. 

109.  Lay  off  on  the  line  E(l  +i  +  lf)  F(4  +  !$-!)  a  point  D,  2"  from  E. 

110.  Lay  off  on  theline  K(2  +  2-i)L(2  +  $-l)  a  point  N,  H"  from  L. 

111.  Lay  off  on  the  line  A(2~i+f)B(2-lf  +  i)  a  point  C,  3"  from  B. 

112.  Through  D(l+l-l)   draw  a  line  2$"  long,  parallel  to  E(3-$  +  l) 


113.  M(2,  0-1)  N(3$,  0-2J)  is  the  diagonal  of  the  base  of  a  right  square 
pyramid,  2$"  high,     (a)  Find  the  length  of  one  of  its  slanting  edges. 
(b)  Find  the  length  of  a  line  running  from  a  point  If"  high  on  one 
edge  to  a  point  $"  high  on  an  adjacent  edge. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES      37 

114.  A  cable  runs  through  a  factory  building  from  A(1^+2J  — ^)  to  B(4^  — 
£  —  2)  in  the  floor  below. 

(a)  What  is  the  length  of  the  cable?     Scale  \"  =  1  ft. 

(b)  At  what  point  in  the  floor  (H}  must  the  hole  be  drilled  for  it? 

(c)  A  support  from  the  wall  (F)  is  required  6  ft.  from  the  floor  hole, 
measured  on  the  cable.     How  far  from  the  waJl  and  the  floor  is  the 
point  of  support? 

115.  Scale  1  in.  =  100  ft.     A  rope  conveyor  is  to  be  constructed  from  a  point 
(M50ft.,0  —  160ft.)  to  a  loading  point  on  a  hill  175  ft.  high,  and  225 
ft.  north,  55°  east  of  the  dumping  point.     How  much  rope  must  be 
purchased  for  a  duplex  belt,  allowing  10  ft.  for  slack,  pulleys,  etc.? 

116.  Fig.  25  shows  the  plan  and  elevation  of  a  plain  hip  roof.     Draw  it  to 
any  convenient  scale. 


FIG.  25. 


(a)  Which  is  its  H-and  which  its  F-projection?     What  angle  is  it 
drawn  in?     Draw  its  P-projection. 

(b)  Locate  the  traces  of  the  roof  planes,  C  and  D. 

(c)  What  is  the  length  of  the  hip  rafter  AB? 

(d)  Locate  a  point  in  the  roof  C,  3  ft.  below  the  ridge,  and  10  ft.  to 
the  right  of  the  center. 

(e)  Locate  a  point  on  the  rafter  EF,  3  ft.  from  F. 

117.  Scale  1  in. -20  ft.  A  60-ft.  stack,  3  ft.  in  diameter,  rises  from  A(60 
ft.,  0  —  35  ft.),  on  the  top  of  a  flat  roof  (H}.  Five  guy-wires  are 
attached  to  the  stack  at  40  ft.  above  the  roof,  and  are  anchored  at 
points  on  the  roof,  respectively,  B(30  ft.,  0-10  ft.),  C(85  ft.,  0,  - 
7  ft.  6  in.),  D(90  ft.,  0-47  ft.  6  in.),  E(70  ft.,  0-67  ft.  6  in.),  and 
F(25  ft.,  0  —  52  ft.  6  in.).  Draw  the  stack  and  guys,  and  measure  the 
true  length  of  the  guys. 

55.  Problem   14. — To   measure   the   angle   between   any   two 
intersecting  lines. 


38 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Two  intersecting  lines  lie  in  one  plane,  and,  if  that  plane  be 
folded  so  as  to  coincide  with  H  or  V,  the  revolved  projections 
will  show  the  true  relation  of  the  lines. 

Analysis. — 1.  Pass  a  plane  through  the  two  lines. 

Note. — One  trace  only  need  be  obtained. 

2.  Revolve  the  plane  about  its  H-tr&ce  into  H}  or  about  its 
F-trace  into  V. 

Note. — Revolving  the  various  points  in  a  plane  about  one  of  its  traces  is 
equivalent  to  revolving  the  plane. 

3.  The  points  are  now  shown  in  their  true  relation  to  each 
other,  and  the  angle  is  in  its  true  size. 

Construction. — Let  it  be  required  to  measure  the  angle  between 
the  lines  AB  (Fig.  26)  and  CD,  which  intersect  at  O. ' 


FIG.  26. 

1.  Find  the  piercing  points,  m  and  ri,  of  the  lines,  and  draw 
the  //-trace,  tt,  through  them. 

2.  Revolve  O  (by  Problem  9)  about  tt  into  //  to  the  point  or 

3.  Connect  m  and  n  with  ot,  and  the  revolved  positions  of 
AB  and  CD  are  found. 

4.  The  required  angle  is  then  either  nojdj  or  nojin. 

Proof. — In  this  construction,  operation  (3)  may  be  explained, 
as  follows:  M  and  N  are  points  in  CD  and  AB  respectively,  and 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       39 

0  is  in  both.  Hence  if  N,  O,  and  M  are  revolved,  both  lines  are 
revolved,  because  two  points  in  any  line  determine  that  line 
fully.  M  and  N,  being  in  the  axis,  will  not  move  in  the  revolu- 
tion (see  Art.  51,  Note  1),  and  so,  if  they  are  connected  with  the 
revolved  position  of  0,  we  shall  have  both  lines  revolved.  The 
revolved  position  of  any  point  in  either  line  may  be  found  by 
running  a  perpendicular  to  the  trace  through  the  H-projection 
of  the  point,  until  it  intersects  the  revolved  position  of  the  line. 
The  points  bx  and  dt  in  Fig.  26  are  located  in  this  manner. 

Notes. — 1.  Either  angle  between  the  lines  may  be  considered  the  angle 
ta  be  measured,  but  it  is  customary  to  choose  the  acute  angle,  unless  other- 
wise specified. 

2.  It  will  sometimes  happen  that  the  piercing  points  .of  the  lines,  or  some 
of  them,  will  be  inaccessible.  Under  such  conditions,  one  or  more  auxiliary 
lines  may  be  drawn  through  points  in  the  given  lines.  These  auxiliaries 
will  lie  in  the  plane,  and  may  be  so  drawn  as  to  determine  the  desired  trace. 


FIG.  27. 

56.  Problem  15. — To  measure  an  angle  when  one  side  is 
parallel  to  H. 

Analysis. — 1.  Revolve  the  oblique  line  about  the  parallel  line 
as  an  axis,  until  it  is  also  parallel  to  H.  This  is  a  modification 
of  Problem  11. 

2.  Both  lines  being  parallel  to  H,  the  angle  is  projected  on 
H  in  its  true  size.  (Theorem  XXII,  Art.  24.) 

Construction. — ABC  (Fig.  27)  is  the  required  angle,  with 
AB  parallel  to  H. 


40 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


1.  Revolve  C  about  AB  as  an  axis  into  the  same  horizontal 
plane  in  which  AB  lies.     To  do  this,  run  the  path  of  revolution 
through  c,  and  construct  the  right  triangle  for  the  radius,  by 
using  c'x'  and  cy  as  the  respective  sides.     The  hypothenuse  will 
be  the  distance  yc1;  laid  off  in  either  direction  from  ab. 

2.  The  angle  abc1;  will  be  the  true  angle  between  the  lines. 

Notes. — 1.  A  problem  like  this  requires  no  GL. 

2.  This  problem  can  be  done  in  the  same  manner  if  one  of  the  lines  is 
parallel  to  V  or  P. 

3.  This  problem  can  be  done  exactly  as  in  Problem  14,  but  the  method 
here  given  is  an  easier  solution  for  this  situation  of  the  lines. 

4.  It  may  help  the  student  to  transfer  GL  to  a'b',  thus  putting  AB  into 
//,  without  changing  the  relation  of  the  lines. 


FIG.  28. 

57.  Problem  16. — To  measure  an  angle,  when  one  side  is 
perpendicular  to  H. 

Analysis. — Revolve  the  oblique  line  about  the  perpendicular 
line  as  an  axis,  until  it  is  parallel  to  V. 

Construction. — Use  the  method  employed  in  solving  Problems 
10  and  11  (see  Fig.  28).  In  this  figure  AB  is  the  perpendicular  to 
H,  CB  the  oblique  line,  and  ABC  the  angle  to  be  measured. 
Rotating  BC  until  be  is  parallel  to  GL,  we  obtain  the  angle 
a'b'c'j,  which  is  the  required  angle. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES      41 

58.  Problem  17. — To  find  the  projection  of  the  bisector  of  any 
angle. 

Analysis. — 1.  Revolve  the  lines  to  show  the  true  angle,  by 
the  method  shown  in  any  of  the  three  preceding  problems,  as 
the  case  may  require. 

2.  Bisect  the  angle  in  its  revolved  position. 

3.  Revolve  any  point  of  the  bisector  back  into  the  original 
plane  of  the  lines. 

4.  Connect  this  point  with  the  vertex. 


FIG.  29. 


Construction. — It  is  required  to  find  the  projections  of  the 
bisector  of  the  angle  ABC,  Fig.  29. 

1.  Obtain  the  trace  tt. 

2.  Revolve  B  to  b1?  and  C  to  c1?  and  connect  a,  b1;  and  cl  to 
obtain  the  true  angle. 

3.  Draw  the  bisector,  djbj,  of  the  angle  ab^. 

4.  Draw  the  auxiliary  line  act,  intersecting  bidl  in  dr 

5.  Revolve  act  back  to  its  original  position,  ac,a'c'. 

6.  Revolve  dt  back  to  d  on  ac. 


42  PRACTICAL  DESCRIPTIVE  GEOMETRY 

7.  Project  d  to  d'  on  a'c'. 

8.  Draw  b'd'  and  bd,  the  required  projections  of  the  bisector. 

Note.  —  When  the  angle  to  be  bisected  is  included  between  the  piercing 
points  and  the  vertex,  as  is  the  angle  ABM  in  Fig.  29,  the  auxiliary  line  is 
unnecessary.  This  is  so,  because  the  bisector  will  intersect  the  trace,  and 
the  return  revolution  can  be  effected  without  difficulty.  Give  the  reason. 

Why  do  not  the  projections  of  the  bisector  of  an  angle  bisect 
the  corresponding  projections  of  the  angle? 

EXERCISES 

59.  Measure  the  angles  included  between  the  following  lines. 

118.  A(2  +  £-l)  B(3  +  H-2i)  and 

119.  D(lf  +  f-2J)  E(4  +  l-|-l)  and 

120.  G(l+2-2)  H(4  +  1-H)  and 

121.  M(2  +  2-2)  N(2  +  l-i)  and  NO(4  +  2-2). 

122.  A(2+2-$)  B(2  +  i-2)  and  BC(4  +  2-£). 

123.  D(l+2-l)  E(2i-l  +  l$)  and  EF(3  +  l  +  2). 

124.  G(l$,  0,  0)  H(3  +  l-li)  and  HK(4f,  0,  0). 

125.  M(2  +  2-2)  N(4+l-$)   and  NO(2  +  !£-!)   by    revolving  into  the 
profile. 

126.  A(l'i  +  2  +  2)  B(3i  +  l  +  l)  and 

127.  D(2-i-4)  E(3i-f-|)  and 

128.  G(2  +  l$-2)  H(4  +  l$^)  and 

129.  M(5  +  lf-lJ)  N(2i  +  l|-2)  and 

130.  A(2  +  2-li)  B(4  +  2-lJ)  and  BC(2  +  1- 

131.  D(l|  +  l-|)  E(4  +  li-|)  and  EF(2  +  J- 
3.32.  Q(2-li  +  l)  H(4-li  +  2)  and  GK(4- 

133.  M(li  +  li-i)  N(lKli~2)  and 

134.  A(2,  0,  0)  B(4-f-2-li)  with  GL. 

Find  the  angle  between  the  traces  of  the  following  planes. 

135.  T(5  +  2) 

136.  S(l  +  2) 

137.  R(l  +  2)  3^(5-2). 

138.  W(l+2)  3^(5-1). 

139.  T(2|  +  2)  2^(1-1),  and  of  U(5  +  2)  3(3-2). 

Measure  the  angles,  and  find  the  true  size  of  the  following  set  of  triangles. 

140.  A(l+i-i)  B(2i  +  2-2£)  0(4  +  2-1). 

141.  E(2-i  +  i)  F(2  +  li-2 

142.  H(2  +  2i-2)  K(3  +  2- 

143.  Measure  the  angles  and  true  size  of  the  parallelogram  A(2+l^  —  1) 


144.  Prove  graphically  that  D  (2  +  1^-2)   E(3J  +  li-i)  F(5  +  i~2)  is  a 
right  angle.     Write  an  explanation. 

145.  Draw  a  line  from  0(21  +  2-1$)  to  intersect  M(l  +  $-f)  N(3i  +  l-l$) 
at  45°. 

146.  Measure  the  angles  between  two  adjacent  edges,  and  between  an  edge 
and  a  base  line  of  the  square  pyramid  given  in  Exercise  113,  Art.  54. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES      43 


147.  Measure   the   angle   that    H(2  +  l-$)    K(4  +  2$  —  1)    makes   with  its 
H  -projection.     Also  that  which  it  makes  with  its  F-projection. 

148.  Draw  the  roof  shown  in  Fig.  25,  and  measure  the  following  angles 
therein  : 

(a)  Hip  rafter  EF  with  hip  rafter  EG. 

(b)  Hip  rafter  EF  with  eave  FG. 

(c)  Hip  rafter  AB  with  eave  AG. 

(d)  Hip  rafter  AB  with  ridge  BE. 

149.  Draw  the  roof  shown  in  Fig.  25  to  a  definite  scale,  and  measure  the  true 
size  of  the  roof  planes  C  and  D. 

Compute  their  areas  from  the  scale. 

What  would  it  cost  to  cover  the  roof  with  material  costing  $1.50  per 

square  yard? 

150.  The  //-trace  of  T(x,  x)  4(2-1$)  makes  75°  with  its  F-trace.     Draw 
the  F-trace. 

151.  A  ray  of  light  from  P(l|  +  l-l$)is  reflected  at  A  in  F  to  B  in  H,  and 
thence  to  0(5  +  2$  —  2£).     Locate  A  and  B. 

152.  Find  the  center  of  the  triangle  C(l  +  l  +  2)  D(2|-2-l)  E(4  +  $  +  !*). 

153.  Draw  a  line    through  O(2  +  l,  x)  in  the  plane  T(l$  +  3) 
making  45°  with  the  //-trace. 

Find  the  projections  of  the  bisectors  of  the  following  angles. 
154. 

155.  D(2  +  2-lf) 

156.  G(l£  +  2-2) 

157.  M(2  +  2  +  2)  N(4  + 

158.  A(2  +  l£-2)  B(3i 

159.  D(2  +  f-l)  E(4  +  f-l)F(3  +  2- 

160.  G(2  +  2-2) 

161.  M(2£  +  li- 

162.  A(2  +  li-2)  B(4,  0,  0)  C(l,  0,  0). 

163.  Between  the  H-  and  F-traces  of  T(l£  +  2)  4(1$-!$). 

164.  Between  the  H-  and  P-traces  of  S(4£  +  l£)   l(4J-2). 

165.  Between  the  H-  and  F-traces  of  R(2  +  2)  4(2  +  1). 

166.  Between  the  H-  and  F-traces  of  X(l  +  2)  3(5-2). 

167.  Between  the  V-  and  P-traces  of  W(  +  l$)  oo(-l). 

168.  D(2  +  li-l£)  E(4  +  $,  0)  and  its  F-projection. 

169.  F(3  +  2-i)  G(3  +  l- 
170. 


60.  Problem  18.  —  To  find  the  shortest  distance  from  a  point 
to  a  line. 

From  Plane  Geometry  we  know  that  the  shortest  distance  from 
a  point  to  a  line  is  on  the  perpendicular  through  the  point  to  the 
line. 

Analysis.  —  1.  Pass  a  plane  through  the  point  and  the  line. 

2.  Revolve  the  point  and  line  about  either  trace  of  the  plane 

4  -  <--•   C.I-./^TTT  4-V>oIv  iriif>   rotation. 


44 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


3.  From  the  revolved  position  of  the  point  erect  a  perj. 
dicular  to  the  revolved  position  of  the  line. 

4.  The  length  of  this  perpendicular  is  the  shortest  distance. 
Construction. — Let  AB  (Fig.  30)  be  the  given  line,  and  C  the 

given  point. 

1.  Draw  the  #-trace  of  the  plane  containing  the  line  and 
point. 

Note. — Either  trace  may  be  used. 


FIG.  30. 

2.  Revolve  AB  to  a^. 

3.  Revolve  C  to  Cj. 

4.  Draw   C1d1   perpendicular   to   a^.     This   is   the   required 
distance. 

61.  Problem  19. — To  find  the  projections  of  any  given  plane 
figure  lying  in  an  oblique  plane. 

This  problem  will  be  found  to  use  the  principles  employed 
in  the  three  preceding  problems,  but  with  the  order  of  operation 
partly  reversed. 

Analysis. — 1.  If  no  point  or  line  of  the  required  figure  is  given, 
assume  such  a  point  or  line  in  the  given  plane. 


PROBLEMS  RELATING.  TO  POINTS,  LINES,  AND  PLANES      45 

2.  Revolve  the  point  or  line  thus  assumed  about  the  H-trace 
into  H,  or  about  the  F-trace  into  V. 

3.  Construct  the  figure  in  its  true  size  on  or  about  the  re- 
volved point  or  line. 

4.  Revolve  the  figure  back  into  the  plane  in  the  original  posi- 
tion of  the  plane. 

62.  Counter-revolution  of  revolved  points  and  lines. 

Return-revolution,  or  counter-revolution,  as  it  is  also  called, 
is  the  only  point  of  this  construction  that  is  new  or  difficult. 
Therefore,  before  taking  up  the  construction,  let  us  consider  it. 
We  have  had  a  specimen  of  it  in  finding  the  projections  of  the 
bisector  of  an  angle,  Problem  17.  Inasmuch  as  this  is,  perhaps, 
the  most  generally  practical  application  of  Descriptive  Geom- 
etry, it  would  be  well  to  do  many  problems  and  exercises  involv- 
ing its  use.  Two  methods  of  return-revolution  will  be  given. 


In  the  revolution  of  any  line  about  the  trace  of  its  plane,  two 
facts  should  be  borne  in  mind;  The  piercing  point  of  the  line  does 
not  move  in  revolution,  because,  as  we  have  seen,  it  is  in  the 
trace;  that  is,  its  axis.  Also,  if  the  line  is  parallel  to  the  trace, 
it  will  be  parallel  when  revolved,  and  the  converse  of  this  is  true. 
With  these  two  rules,  any  problem  in  counter-revolution  may 
be  worked. 

Let  M  (Fig.  31)  be  a  known  point  in  the  plane  T.  Revolve 
M  about  the  #-trace  to  m^  Through  nij  draw  any  line  m^. 


46  PRACTICAL  DESCRIPTIVE  GEOMETRY 

It  is  desired  to  counter-revolve  the  line  m^  into  the  plane  T. 

It  is  obvious  that  nij  will  return  to  mm';  that  is,  where  it 
came  from. 

If  m^  is  parallel  to  Tt,  it  will  be  parallel  when  it  is  re- 
turned to  T,  and  its  ^/-projection  can  be  drawn  through  m 
parallel  to  the  //"-trace.  The  point  nA  will  of  course  return  in  a 
circle  perpendicular  to  the  axis. 

If  m^  is  not  parallel  to  Tt,  mn  will  not  be  parallel  to  Tt,  and 
both  will  therefore  have  to  intersect  Tt.  From  what  we  already 
know  of  revolution,  we  know  these  intersections  must  be  in  one 
point. 

Extend  m^  to  Tt  at  the  point  p.  The  line  mtp  will  contain 
the  point  nt,  so  n  will  lie  on  mp.  We  know  the  actual  positions 
in  space  of  M  and  P,  and  as  N  must  lie  in  that  line,  its  projections 
must  be  on  the  projections  mp,  m'p'.  Revolve  nt  on  a  perpen- 
dicular to  the  trace  to  the  line  mp,  and  then  project  to  m'p', 
and  we  have  the  line  MN  revolved  back  into  the  plane. 

This  is  the  more  convenient  method  of  counter-revolution, 
but  some  may  find  the  work  easier  understood  by  the  Similar 
Triangles  Method. 

SIMILAR  TRIANGLES  METHOD 

This  method  depends  on  the  fact  that  all  points  in  a  plane, 
when  rotating  about  one  of  the  traces,  have  for  radii  the  hy- 
pothenuses  of  similar  triangles. 

Referring  to  Fig.  31,  mxx  and  nty  are  the  radii  of  the  revolu- 
tion of  the  points  M  and  N  about  Tt.  They  are,  therefore,  the 
hypothenuses  of  triangles,  whose  bases  and  altitudes  are  respec- 
tively mx,  ny,  m'a,  and  n'b,  and  the  triangles  are  similar.  So, 
to  revolve  a  point,  as  N,  back  to  the  original  position  of  the  point 
in  the  plane,  1".  Draw  the  path  of  revolution  through  i\l  perp^n- 
dicular  to  the  trace,  intersecting  it  at  y. 

2.  The  perpendicular  nty  will  be  the  hypothenuse  of  the  tri- 
angle to  be  formed. 

3.  With  ynx  as  a  hypothenuse,  construct  a  triangle  similar 
to  mzx. 

4.  The  sides  will  be  respectively  equal  to  ny  and  rib. 

5.  Lay  off  ny  on  the  path  of  revolution,  locating  n. 

6.  Project  n  to  n',  locating  it  in  the  plane  by  any  of  the  regular 
methods. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES      47 

Construction. — Let  it  be  required  to  draw  the  projections  of 
a  1-in.  square  in  T  (Fig.  32)  about  the  point  O  (in  T)  as  a  center. 
Let  it  be  stipulated  that  two  sides  of  the  square  be  parallel  to  H. 

1.  Locate  O  in  T  by  Problem  5. 

2.  Revolve  O  about  Tt  into  H  at  ox  by  Problem  9. 

3.  Construct  the  1-in.  square  a^c^  about  ol  as  a  center,  in 
its  actual  size.     As  two  sides  are  specified  parallel  to  H,  we 
draw  cldl  and  a^  parallel  to  Tt.     See  Theorems  XXVII  and 
XXVIII. 


FIG.  32. — A  square  drawn  in  an  oblique  plane. 

4.  Draw  the  auxiliary  line  bjX  through  bt,  O1?  and  cr 

5.  Draw  xo;  prolonging  it  as  much  as  necessary. 

6.  Revolve  bx  and  ct  about  Tt  to  the  points  b  and  c  on  the 
line  xo. 

7.  Draw  ab  and  cd  parallel  to  Tt,  and  complete  the  parallel- 
ogram abed. 

8.  Project  a,  b,  c  and  d  to  their  ^-projections,  locating  them 
in  T,  at  the  points  a/  b',  c'  and  d'. 

This  gives  us  both  projections  of  the  square. 

Notes. — 1.  The  points  might  have  been  rotated  back  into  T  by   the 
similar  triangles  method. 


48  PRACTICAL  DESCRIPTIVE  GEOMETRY 

2.  It  is  sometimes  desirable  to  show  the  F-traee  revolved  about  the 
//-trace  into  H.  It  is  desirable  in  order  to  limit  the  figure  to  one  space  angle. 
By  taking  E  (Fig.  32)  any  point  on  the  F-trace,  and  revolving  it  about  the 
H- trace  into  H  at  ei;  the  revolved  trace  may  be  drawn  through  T  and  er 
Note  that  Te'  and  Te,  are  equal,  and  that  the  revolution  can  be  made  by 
striking  an  arc  with  Te'  as  a  radius  until  it  intersects  the  perpendicular 
from  e  to  the  trace  at  er 

63.  EXERCISES 

Find  the  shortest  distance  from  the  following  points  to  the  following  lines: 

171.  A(2^  +  2-2)  to  B(2  +  i-l)  C(4  +  2,  0). 

172.  D(3  +  li-2)  to  E(2  +  l-i)  F(4 +  !-£). 

173.  G(3  +  li-2)  toGL. 

174.  H(3-l-l)  to  K(4  +  2  +  2)  L(4  +  l  +  l). 

175.  M(2  +  l,  x)  in  T(2  +  3)  4^(2-2)  to  the  //-trace,  and  to  the  F-trace. 

176.  N(2i,  x,  0)  on  the  F-trace  of  T(2  +  3)  4$  (2 -2)  to  the  //-trace. 

177.  A  3-in.  steam  pipe  runs  through  a  basement  from  A(3  ft.,  0  —  9  ft.)  to 
B(15  ft.   +  7  ft.  6  in.,  0).     It  is  desired  to  run  the  shortest  possible 
1^-in.  pipe  from  a  wall  connection  at  C(5  ft.  3  in.,  +  7  ft.  6  in.,  0)  to  the 
main  pipe.     Allowing  3  in.  for  connections,  what  will  be  the  length  of 
the  smaller  pipe?     Scale  1  in.  =10  ft. 

Draw  the  projections  of  the  following  plane  figures  according  to  the  condi- 
tions given. 

178.  A  li-m.  square  in  T(l+3)  4(1-2),  one  side  [|  to  H,  and  £  in.  from 
the  //-trace. 

179.  A  li-m.  square  in  T(l+3)  4(1-2),  center  at  O(2  +  l,  x),  its  diagonal 
[jtoF. 

180.  A  right  triangle,  base  1J  in.,  altitude  f  in.  in  T(l  +3)  4(1-2).     Center 
at  O(2  +  l,  x)  in  T. 

181.  An   equilateral  triangle,  l^-in.  sides,  in  T(l+3)  4(1-2).     Center  at 
O(2  +  l,  x)inT. 

182.  A  regular  hexagon   of   1-in.    sides,   in   T(l  +  3)    4(1-2).     Center  at 
O(2  +  l,  x)  in  T. 

183.  A  regular  pentagon    of    1-in.    sides   in    T(l+3)  4(1-2).     Center  at 
0(2  +  1,  x)  in  T. 

184.  A  li-in.  circle  in  T(l+3)  4(1-2).     Center  at  O(2  +  l,  x)  in  T. 

Note. — Both  projections  will  be  ellipses,  and  each  major  axis  will  be 
parallel  to  the  respective  trace.  Find  the  minor  axis  of  each  ellipse,  which 
will  be  the  diameter  of  the  circle  perpendicular  to  the  respective  trace. 
Having  these  two  dimensions  of  each  ellipse,  draw  the  curves  by  any  of  the 
usual  methods,  preferably  the  "trammel  method."  Of  course  there  can 
be  no  objection  to  dividing  the  circle  into  any  number  of  parts,  and  counter- 
revolving  the  points  thus  obtained,  but  it  is  more  laborious.  It  should,  of 
course,  be  noted  that  the  two  minor  axes  are  not  projections  of  the  same 
line. 

185.  Inscribe  a  circle  in  the  triangle,  Exercise  181. 

186.  Draw  a  1-in.  circle  in  the  center  of  the  hexagon  in  Exercise  182. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES      49 

187.  A  1-in.  square  in  S(  +  2)oo(  — 1$),  one  side  ||  to  H  and  f  in.  from  the 
//-trace. 

188.  A  IHn.  square  in  S(  +  2)oo(-l$),  the  diagonal  \\  to  H,  center  at 
M(3  +  £,  x)  in  S. 

189.  A  right  triangle,  base  li  in.,  altitude  £  in,  in  S(  +  1$)  oo(  -  If ).     Draw 
its  hypothenuse  ||  to  //,  $  in.  from  the  //-trace. 

190.  An  equilateral  triangle,  IHn.  sides,  in  S(  +  2)oo(  —  1$),  whose  base  is 
in  the  line,  which  is  the  locus  of  all  points  in  S  that  are  $  in.  above  H. 

191.  A  regular  hexagon,  1$  in.  across  the  flats,  in  S(  +  2)oo(  — 1$).     Center 
at  K(3  +  £,  x)  in  S. 

192.  A  regular  octagon,  f-in.  sides,  in  S(  +  2)  oo(  — 1$).     Center  at  O  (3  + f,  x) 
in  S. 

193.  A  l$-in.,  long  diameter,  regular  hexagon  with  a  f-in.  circle  in  the  center 
in  S(  +  2)oo(-l$).     Center  at  M(3  +  l,  x)  in  S. 

194.  Inscribe  a  circle  in  the  triangle  in  Exercise  190. 

195.  A  1-in.  square  in  R(l  +  2)3  (5  -2),  ||  to  H,  center  at  A(2$  +  l$,  x)  in  R. 

196.  A  IHn.  square  in  R(l  +  2)  3(5-2),  its  diagonal  ||  to  F,  1  in.  from  the 
F-trace. 

197.  A  right  triangle,  whose  base  is  1£  in.,  altitude  f  in.,  in  R(l  +  2)  3(5  —  2). 
Draw  the  hypothenuse  1 1  to  H ,  f  in.  from  the  //-trace. 

198.  An  equilateral  triangle  in  R(l  +  2)  3(5  —  2),  whose  base  is  the  line 
A(2  +  2,x)  B(3  +  l,  y)  in  R. 

199.  A  regular  hexagon,  1$  in.  long  diameter,  in  R(l  +  2)  3(5  —  2),  its  center 
at  O(2$  +  l$,  x)  in  R. 

200.  A  l$-in.  circle  in  R(l  +  2)  3(5-2),  center  at  M(2$  +  l$,x)  in  R. 

201.  A  2-in.  octagon  in  R(l  +  2)  3(5-2),  two  sides  ||  to  H,  center  at  M(2$  + 
1$,  x)  in  R. 

202.  A  l$-in.  square  in  Q(3  +  3)  3(5-2),  diagonal  \\  to  H,  center  1$  in. 
above  H  and  1  in.  from  F. 

203.  A  l$-in.    square   in  Q(3  +  3)  3(5-2),  two    sides    ||    to  H,   center  at 
M(4  +  l,  x)  inQ. 

204.  A  right  triangle,  base  U  in.,  altitude  1  in.,  in  Q(3  +  3)  3(5-2),  hypoth- 
enuse ||  to  F,  $  in.  from  F. 

205.  An  equilateral  triangle  in  Q(3  +  3)  3(5-2),  its  base  A  (4  +  1$,  x)  B(5  + 
$,  x)  in  Q. 

206.  A  regular  hexagon,  1-in.  sides,  in  Q(3  +  3)  3(5  -  2),  center  at  M(4  + 1,  x) 
inQ. 

207.  A  regular  hexagon,  l$-in.  short  diameter,  in  Q(3  +  3)  3(5-2),  with  its 
diagonal  on  the  line  E(3  +  l,  0)  F(5,  0-2).     Draw  a  1-in.  circle  in  its 
center. 

208.  Inscribe  a  circle  in  the  triangle  in  Exercise  205. 

209.  Circumscribe  a  circle  about  the  hexagon  in  Exercise  206. 

210.  A  1-in.  square  in  W(l  +3)  3(5-1$),  with  one  side  in  the  line  C(3$  + 1,  x) 
D(4$  +  $,  x)  in  W. 

211.  A  square  in  W(l  +  3)  3(5-1$),  whose  diagonal  is  E(3$  +  H,  x)  F(4$  + 
f ,  y)  in  W. 

212.  A  IHn.  square  in  W(l+3)  3(5-1$)  with  one  corner  touching  //,  and 
a  side  making  30°  with  the  //-trace. 

213.  A  IHn.  square  in  W(l+3)  3(5-1$),  having  two  sides  parallel  to  H, 

4 


50  PRACTICAL  DESCRIPTIVE  GEOMETRY 

one  of  them  |  in.  from  the  //-trace,  the  entire  square  being  within  the 
first  angle. 

214.  A  l$-in.  equilateral  triangle  in  W(l  +  3)  3(5  - 1$),  center  at  A(4$  +  f ,  x), 
with  one  corner  touching  H. 

215.  Inscribe  a  circle  in  the  triangle  in  Exercise  214. 

216.  A  l$-in.  hexagon  in  W(l+3)  3(5-1$),  with  two  sides  parallel  to  H. 
Center  at  M(4  +  f,  x)  in  W. 

217.  An  octagon,  sides  fin.,  in  W(l+3)  3(5  —  1$),  whose  center  is  M  (4+1,  x) 
in  W. 

218.  A  circle  in  W(l  +  3)   3(5  —  1$),  one  of  whose  diameters  is  the  line 
K(3$  +  l$,  x)  L(4$  +  f,  y)  in  W. 

219.  The  roof  of  a  tower  is  a  hexagonal  pyramid,  whose  base  is  12  ft.,  long 
diameter,  and  whose  altitude  is  15  ft.     In  each  of  the  three  front  sides 
are  rectangular  windows  3  ft.  by  5  ft.,  each  3  ft.  from  the  base  of  the 
tower.     Draw  the  projections.     Scale  %  in.  =  1  ft. 

64.  Problem  20. — To  find  the  line  of  intersection  of  two  planes. 
Axioms. — 1.  The  intersection  of  two  planes  is  a  straight  line.. 
2.  Two  points  determine  a  straight  line. 

Analysis. — 1.  Find  both  projections  of  the  point  where  the 
F-traces  of  the  given  planes  intersect. 

2.  Find  both  projections  of  the  point  where  the  ^/-traces 
intersect. 

3.  Join  the  respective  projections  of  these  two  points,  which 
are  common  to  both  planes. 

4.  The  line  thus  drawn  is  common  to  both  planes,  and  therefore 
their  intersection. 

Note. — The  //-projection  of  any  F-trace  is  in  GL,  and  therefore  the  //-pro- 
jection of  any  point  in  the  F-trace  is  in  GL. 

Proof. — The  F-traces,  being  each  a  line  in  one  of  the  given 
planes,  intersect  in  a  point  common  to  both  planes,  and  therefore 
in  the  line  of  intersection.  Similarly  for  the  intersection  of  the 
//-traces.  Having  found  these  two  points,  the  line  of  inter- 
section is  determined. 

Construction. — Left  to  the  student. 

65.  Problem  21.  To  solve  the  foregoing  problem  when  the 
traces  do  not  intersect  within  the  limits  of  the  problem,  or  not 
at  all. 

Analysis. — 1.  If  one  pair  of  traces  intersect,  locate  their  point 
of  intersection. 

2,  Draw  an  auxiliary  plane  that  will  intersect  both  given 
planes.  The  lines  of  intersection  of  the  auxiliary  and  given  planes 
will  intersect  in  a  point  on  the  required  line  of  intersection. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       51 

3.  If  neither  pair  of  traces  intersect,  draw  a  second  auxiliary 
plane  which  will  determine  another  point  on  the  required  line, 
thus  determining  the  'line  of  intersection. 

Construction. — For  most  oblique  planes,  such  as  the  two  shown 
in  Fig.  33,  it  is  quickest  and  most  convenient  to  use  planes 
parallel  to  H  or  V  as  auxiliaries.  Such  planes  have  one  trace 
only,  and  that  is  parallel  to  GL.  Such  planes  intersect  oblique 


r  - 


planes  in  lines  parallel  to  their  traces;  that  is,  a  plane  parallel 
to  V  will  intersect  an  oblique  plane  in  a  line  parallel  to  its 
F-trace,  etc.  Let  the  student  give  the  reason  from  Solid 
Geometry. 

1.  Draw  q'q',  the  7-trace  of  Q,  parallel  to  H.     The  lines  cut 
by  Q  from  T  and  S  are  CM  and  DM. 

2.  Locate  m  and  m',  their  intersection. 

3.  I}raw  rr,  the  #-trace  of  R,  parallel  to   V.     The  lines  cut 
by  R  from  T  and  S  are  AO  and  BO. 


52 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


4.  Locate  o  and  o',  their  intersection. 

5.  Draw  o'm'  and  om,  the  required  line  of  intersection. 

66.  Special  Case  i. —  It  sometimes  happens  that  only  one 
point  on  the  line  of  intersection  can  be  conveniently  located  by 
the  auxiliary  planes  just  shown.  Fig.  33a  gives  such  a  case. 
In  a  case  like  this  an  auxiliary  plane  parallel  to  one  of  the 
planes  may  be  used  to  determine  the  direction  of  the  line  of 
intersection.  This  is  true,  because  two  parallel  planes  cut  a 
third  plane  in  parallel  lines.  Fig.  33a  shows  two  planes, 
T  and  S,  in  which  an  auxiliary  plane  Q,  parallel  to  S,  is  used. 


Q  intersects  T  in  the  line  XY.  The  //"-traces  of  S  and  T  inter- 
sect in  the  point  M.  Through  M  draw  MN  parallel  to  XY. 
MN  will  be  the  intersection  of  S  and  T. 

Special  Case  2. —  When  both  planes  are  parallel  to  GL. 

In  this  case  the  traces  are  all  parallel,  yet  the  planes  may 
intersect.  The  line  of  intersection  will  be  parallel  to  GL, 
because  both  planes  are.  Hence,  one  point  on  the  line  of  inter- 
section is  all  that  is  necessary  to  determine  the  line. 

Figs.  34  and  35  show  two  different  auxiliary  planes  that  may 
be  used.  In  Fig.  34  a  profile  plane  is  used,  and  the  profile 
traces  yield  the  intersecting  point. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       53 

In  Fig.  35  an  oblique  plane,  Q,  is  used,  which  intersects  T 
in  the  line  CD,  and  S  in  the  line  EF.  The  lines  CD  and  EF 
intersect  in  the  point  G,  through  which  the  line  of  intersection, 
AB,  may  be  drawn. 


/     i 


FIG.  34. 


-S 


s- 


-  S 


FIG.  35. 


67.  Note. — The  foregoing  solutions  give  an  idea  of  the  multitude  of 
resources  open  to  the  student  in  the  solving  of  varieties  of  problems.  Some- 
times one  method  or  auxiliary  plane  will  be  useful,  which  would  be  awkward 
under  other  conditions.  No  stereotyped  method  of  performing  these 
solutions  can  be  given,  for  each  problem  will  bring  up  conditions,  where  there 
is  a  preference  of  methods.  The  analyses  are  exactly  alike,  and  they  must 
be  learned,  but  not  memorized. 


54  PRACTICAL  DESCRIPTIVE  GEOMETRY 

EXERCISES 

68.  Find  the  intersections  of  the  following  planes. 

220.  T(2  +  2)  4|(2-24)  and  S(5  +  l|)  3(5-2.). 

221.  R(14  +  3)  4(1-1)  and  Q(34  +  3)  3(5-1). 

222.  X(l+3)  4(5-2)  and  Y(5  +  3)  2(1-3). 

223.  W(2  +  3)  1(5-3)  and  Z(l+3)  4(4-3). 

224.  T(l+24)  44(1-3)  and  S(2  +  3)  2(5-2). 

225.  R(5  +  24)  2(5-3)  and  M,  parallel  to  H,  1  in.  above.     Also  R  with  Q, 
parallel  to  V,  1  in.  behind. 

226.  T  (  +  ll)oo(-f)  and  S(2  +  3)  4(2-3). 

227.  X(14  +  24)  3(5-2)  and  Y(  +  2)oo(-ll). 

228.  W(5  +  3)   2(5-2)   and  a  plane  containing  the  point  A(3J  +  24~24) 
and  GL. 

229.  R(2  +  24)  4(2-3)  and  Q(-|)oo(-2). 

230.  T(l  +  2)  4(54-3)  and  S(5  +  24)  24(1-2). 

231.  X(14  +  3)  4(54-3)  and  Y(44  +  3)  2(44-3). 

232.  W(2  +  3)  4(2-3)  and  Z(34  +  3)  5K4-3). 

233.  Q(4  +  3)  2(4-2)  and  R(2  +  3)  21(54-2). 

234.  T(l  +  l)  4(1-14)  and  S(2  +  3)  44(54-1). 

235.  X(l  +  3)  3(5-3)  and  Y(5  +  2)  2(4-1). 

236.  W(l+3)  (41  +  2)  (1-lf)  (3-ll).andZ(l  +  21)  (6  +  f)  (3|-3)  (51-|). 

237.  Q(l+2)  34(1-3)  and  R(4J  +  2)  34(5-1). 

238.  S(5  +  2i)  31(14-21)  andT(l+2)  31(54-2).. 

239.  W(  +  2)oo(-4)  andX(  +  ll)c»(-ll). 

240.  Y(  +  14)'oo(-2)  andZ(-4)oo(-f). 

241.  Q(-f)oo(-4)  and  R(-14)oo(  +  l). 

242.  S(  +  2)oo(  +  l)  and  T(  +  ll)oo(  +  2J). 

243.  W(  +  2)oo(-f)  andX(-4)c»(  +  2). 

244.  Draw  a  line  AB  parallel  to  S(2  +  2)  44(2-24)  and  T(5  +  14)  2f  (5-24) 
through  the  point  A(4  +  2  — 2). 

245.  Find  the  point  of  intersection  of  the  three  planes  X(4£  +  2)  1(44  —  3), 
Y(2  +  3)  5(3-3)  and  Z(  +  J)oo(-lJ). 

246.  Find  the  point  of  intersection  of  the  three  planes,  T(3  +  3)   1(4-3), 
S(l  +  l)  5(6-1)  and  R(2-2)  4(6  +  2). 

247.  Find  the  projections  of  the  tetrahedron  included  between  the  planes 
X(l+2)  2(5-2),  Y(5  +  2)  44(1-2),  Z(  +  24)oo(-2)  and  H. 

248.  Develop  the  surface  of  the  tetrahedron  in  Exercise  247. 

249.  Find  the  intersection  of  the  plane  T(l  - 1)  54(3  —  3)  and  the  tetrahedron 
in  Exercise  247. 

250.  Draw  the  projections  of  an  equilateral  triangular  prism,  of  J-in.  sides, 
whose  ridge  is  the  line  A(l  +1  —  14)  B(34  + 1  —  4)>  and  of  a  cube  resting 
on  H,  whose  base  diagonal  is  the  line  0(2  +  0-1)  D(34,  0-2).     Find 
the  line  of  intersection  of  cube  and  prism  without  finding  the  traces  of 
their  bounding  planes. 

251.  Draw  the  projections  of  a  pyramid  A(2,  0-  if)  B(3,  0  -f)  0(31,  0-  If) 
D(2f,  0-2f)  E(2|  +  14-14)  and  the  triangular  prism  of  f-in.  sides, 
one  side  parallel  to  H,  whose  center  line  is  K(lf  +  |-l)L(4^  +  f-14). 
Draw  their  line  of  intersection. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       55 

69.  Problem  22. — To  find  the  piercing  point  of  a  line  and 
plane. 

Analysis. — 1.  Pass  a  plane  through  the  given  line. 

2.  Find  the  line  of  intersection  between  the  given  plane  and 
the  auxiliary  plane. 

3.  Where  this  intersection  crosses  the  given  line  is  the  required 
piercing  point. 

Note. — Memorize  this  analysis.     It  is  one  of  the  most  important  in  the 
work. 

Proof. — Since   the    auxiliary   plane  contains  both  the  given 
line  and  the  intersection  with  the  given  plane,  the  two  lines 


FIG.  36. 


must  intersect,  or  be  parallel.  If  they  are  parallel,  the  given 
line  is  parallel  to  the  given  plane,  and  there  is  no  piercing  point. 

Construction. — Let  AB  (Fig.  36)  be  the  given  line,  and  T  be 
the  given  plane. 

1.  Pass  a  plane  (Q)  through  AB. 

Note. — In  nearly  all  cases  the  simplest  plane  to  draw  is  a  projecting  plane. 
In  Fig.  36  the  plane  q'Qq  is  the  //"-projecting  plane  of  AB.  Any  plane  con- 
taining the  line  may  be  used,  but  the  projecting  plane  is  almost  always 
the  easiest. 


56 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


2.  Find  MN,  the  intersection  of  T  and  Q.  ' 

3.  MN  crosses  AB  at  C,  and  therefore  C  is  the  required  piercing 
point  of  AB  and  T. 

70.  Problem  23. — To  find  the  piercing  point  of  a  line  and 
plane  without  obtaining  the  traces  of  the  plane. 

Analysis. — Let  the  plane  be  determined  by  two  intersecting, 
or  two  parallel  lines. 

1.  Pass  a  projecting  plane  through  the  line. 

2.  Find  their  piercing  points  with  this  plane. 

Note. — This  operation  requires  no  auxiliary  planes  or  lines,  as  will  be  seen 
in  the  construction. 

3.  Join  these  piercing  points  by  a  line. 

4.  Where  the  line,  or  its  extension,  crosses  the  given  line  will 
be  the  required  piercing  point. 


FIG.  37. 

Construction. — Let  AB,  Fig.  37,  be  the  given  line.     Required 
to  find  its  piercing  point  with  the  plane  of  MN  and  OP. 

1.  Draw  the  #-trace,  Qq,  of  the  ^-projecting  plane  of  AB. 

Note. — One  trace  only  is  necessary. 

2.  The  points  in  which  this  auxiliary  plane  crosses  the  pro- 
jections op  and  mn  will  be  the  H -projections,  c  and  d,  of  the 
piercing  points  of  OP  and  MN  with  Q. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       57 

3.  Erect  projectors  from  c  and  d  to  c'  and  d'  on  m'n'  and  o'p', 
and  join  c'd'. 

4.  Where  c'd'  crosses  a'b'   at  x'  is  the   7-projection  of  the 
required  piercing  point. 

5.  Project  x'  to  x  on  ab  for  the  required  ^-projection. 

Note. — The  profile  plane  cannot  often  be  used  in  these  problems,  although 
many  students  try  to  use  it  on  every  occasion.  It  is  only  necessary  or 
available  when  dealing  with  a  line  whose  projections  are  perpendicular  to 
GL,  and  sometimes  when  dealing  with  a  plane  parallel  to  GL. 

EXERCISES 

71.  Find  the  piercing  points  of  the  following  lines  and  planes. 

252.  A(2  +  |-l)  B(4  +  3-2)  and  T(l+3)  4$(l-2$). 

253.  C(5  +  2-$)  D(3$  +  $-2)  and  S(2  +  2)  5(2-3). 

254.  E(3  +  2-2)  F(3  +  l-$)  and  R(l+3)  4$(1  -2$). 

255.  G(2  +  l-l)  K(4$  +  l-2$)  and  Q(4  +  3)  3(4-2$). 

256.  M(2  +  l±-f)  N  (4 +  11-1)  and  W(l  +  3)  4$(1  -2$). 

257.  O(l$  +  2-2)  P(4  +  $-$)  and  Y(2  +  3)  2(5-2$). 

258.  A(3  +  $-2)  B(5  +  2-$)  and  T(l  +  3)  2$(5-li). 

259.  0(3  +  1-2)  D(4$  +  2-l)  and  S(5i  +  l$)  3(f-l$). 

260.  E(5  +  l$-lf)  F(4  +  $-H)  and  R(5  +  3)  2(5-1$). 

261.  G(l$  +  2-2)  K(3$  +  l-$)  and  Q(  +  H)oo(-f). 

262.  M(3  +  2-2)  N(3  +  l-$)  and  W(  +  l$)oo(-l). 

263.  A(2 +  1$-|)  B(4  +  l$-f)  and  0(1$ +  !$  +  !$)  D(3$-lf-lf)  and  the 
plane  X(5  +  2)  3(1—2).     Find  the  distance  between  the  piercing  points 
of  AB  and  CD  with  X. 

264.  E(2  +  $  +  $)  F(4  +  2  +  2)  and  Y(l+3)  4$(l-2). 

265.  M(2$  +  2$-2)  N(4  +  $-$)  andZ(l  +  2)  3(5-U). 

Find  the  piercing  points  of  the  following  lines  and  the  planes  of  the  given 
lines,  without  finding  the  traces  of  the  planes. 

266.  A<4  +  1-H)  B(5$  +  2-lf).with  the  plane  of  C(H  +  i-lf)  D(2$  +  l 
-i)  andDE(3-|-f). 

267.  G(lf  +  3i-2$)    H(2|  +  ll-l|)    \vith    the    plane    of    K(l$  +  l£-$) 
L(2$  +  l-l)  and  LO(2  +  i-l$). 

268.  A(3  +  2-2)  B(3  +  $-l)  with  the  plane  of  C(3$  +  2-$)  D(3$  +  $-2) 
and  E(4$  +  2-$)  F(4$  +  $-2).r 

269.  From  the  middle  point,   O,  of  the  line  G(3$  +  J-1)   H(4|  +  l-l|) 
draw  a  line  which  will  intersect  both  K($  +  l$  — 3)  L(l$  +  3-2f)  and 
M(2  +  l$-2$)  N(3  +  2-i). 

270.  Erect  a  square   pyramid  of  2-in.  altitude,  on  H,  having  for  its  base 
diagonal    the  line   A(2|,    0-f)    B(4,    0  —  1$).     Find   where   the   line 
0(4$  +  If  —  1)  D(l$  +  $  — 2)  pierces  the  pyramid,  entering  and  leaving. 

271.  Find  the  shortest  line  that  can  be  drawn  on  the  surface  of  the  pyramid, 
in  Exercise  270,  between  the  two  piercing  points  of  the  line  CD  in  the 
same  Exercise. 

272.  Pass  the  plane  T(l  +  1$)  5(3-3)  through  the  pyramid  in  Exercise  270. 
Crosshatch  the  section,  and  show  it  in  its  true  size  also. 


58  PRACTICAL  DESCRIPTIVE  GEOMETRY 

» 

273.  Develop  the  surface  between  H  and  T  of  the  pyramid  as  found-  in 
Exercise  272.     Make  a  paper  model  of  it. 

274.  Find  the  true  length  of  that  part  of  the  line  A(l  +2-2$)  B(4  +  $  -$) 
included  between  the  planes  T(3  +  2)  1$(3-1$)  and  S  (5 +  3)  3(5-3). 

275.  Find  the  true  length  of  that  part  of  the  line  0(2  +  $-$)  D(4  +  2$-l$) 
included  between  the  planes  X(  +  2J)  oo(  -  If)  and  Y(  +  1±)  oo(  - 1-£). 

72.  Problem  24. — To  draw  a  line  through  a  point  perpendicular 
to  a  plane. 

Analysis. — According  to  Theorem  XXIV,  Art.  25,  a  line  that 
is  perpendicular  to  a  plane  must  have  its  projections  perpendicular 
to  the  respective  traces  of  the  plane;  therefore, 

1.  From  the  projections  of  the  point  draw  projections  of  the 
line  perpendicular  to  the  respective  traces  of  the  plane,  H  to  H , 
and  V  to  V. 

Let  the  student  make  the  construction. 

73.  Problem  25. — To  draw  a  perpendicular  of  any  given  length 
from  a  point  in  a  plane. 

Analysis. — 1.  Erect  a  perpendicular  of  indefinite  length  from 
the  given  point,  according  to  Problem  24. 

2.  Assume  a  point  on  the  perpendicular. 

3.  Revolve  the  definite  line,  thus  drawn,  so  as  to  show  its 
true  length. 

4.  Lay  off  the  required  distance  on  the  true  length  of  the  line 
from  the  given  point. 

5.  Return  the  limiting  point  to  its  proper  position  in  space. 
Compare  this  with  Problem  13. 

Construction. — Let  it  be  required  to  erect  a  perpendicular  1  in. 
long  at  the  point  A  in  the  plane  T,  Fig.  38. 

1.  Draw  ab  and  a'b',  respectively  perpendicular  to  t'T  and  Tt. 

2.  Assume  b'b,-  anywhere  on  the  perpendicular. 

3.  Measure  AB,  true  length  a'b^. 

4.  Lay  off  a'c'j  on  a'b'i,  1  in.  from  a'. 

5.  Return  c\  to  c'  on  a'b'. 

6.  Project  to  c  on  ab. 

AC  is  then  perpendicular  to  T  and  1  in.  long. 

74.  Problem  26. — To  draw  the  projections  of  a  solid  with  its 
base  on  a  given  oblique  plane. 

Analysis. — 1.  Lay  out  the  base  according  to  Problem  19. 

If  it  be  a  prism, 

2.  Erect  perpendiculars  to  the  plane  of  the  base  of  the  specified 
altitude  from  each  of  the  corners  of  the  base,  according  to 
Problem  25. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       59 
V 


FIG.  39. 


60  PRACTICAL  DESCRIPTIVE  GEOMETRY 

3.  Connect  the  ends  of  these  perpendiculars  in  order,  thus 
drawing  the  opposite  base. 
If  it  be  a  pyramid, 

2.  Erect   a  perpendicular  to    the  plane   of  the  base   of  the 
specified  altitude  at  the  center  of  the  base. 

3.  Connect  the  end  of  the  perpendicular,   the   apex  of   the 
pyramid,  with  the  corners  of  the  base. 

Note.  —  In  the  case  of  curved-surfaced  solids,  the  procedure  is  the  same, 
but  only  the  outside  elements,  those  tangent  to  the  curves,  are  drawn. 

Construction.  —  Let  it  be  required  to  draw  a  square  pyramid, 
of  1-in.  base  and  2-in.  altitude,  with  the  center  of  the  base  at  O 
in  the  plane  T,  Fig.  39. 

1.  Revolve  O'to  ot  about  the  //"-trace  as  an  axis. 

Note.  —  Under  certain  conditions  it  will  be  preferable  to  rotate  about  the 
V-  trace,  and  under  others  it  will  be  preferable  to  rotate  about  the  P-trace. 
The  student  will  have  to  judge  from  the  layout  of  his  problem. 

2.  Construct  the  square  in  its  true  size,  a^c^,  about  ox 
as  a  center. 

3.  Counter-revolve  the  square  into  T  (Problem  19). 

4.  Erect  the  perpendicular  OP  at  O. 

5.  Revolve  OP  parallel  to  V  (or  H). 

6.  Lay  off  2  in.  on  o'p'^  at  the  point  m^ 

7.  Counter-revolve  m1  to  m^m  in  o'p'>  op. 

8.  Connect   M  with  A,  B,  C  and  D,  and  make  the  invisible 
lines  dotted. 

75.  EXERCISES 


276.  Through  A(4  +  l£-l)   draw  a  perpendicular  to  the  plane  T(l  +  2) 
3(3  —  2),  and  find  where  it  pierces  it. 

277.  Through  B(2  +  2-l^)  draw  a  perpendicular  to  the  plane  S(4|  +  3) 
2(4|  —  2),  and  find  where  it  pierces  it. 

278.  Through  C(3  +  2-l)   draw  a  perpendicular  to  the  plane  R(f 
2(5  —  3),  and  find  where  it  pierces  it. 

279.  Through  D(3  +  2-3)  draw  a  perpendicular  to  the  plane 
(  —  1),  and  find  where  it  pierces  it. 

280.  Through  E  (3  +  1^-2)   draw  a  perpendicular  to  the  plane  X(l+3) 
5(1  +  1£),  and  find  where  it  pierces  it. 

281.  A  tower,  which  is  a  hexagonal  pyramid  of  16  ft.  base,  and  20  ft. 
altitude,  has  the  center  of  its  base  at  A(24  ft.,  0  —  16  ft.).     From  a 
point  O(38  ft.  +11  ft.  —8  ft.)  a  telephone  wire  is  to  be  run  into  the 
tower  perpendicularly  to  the  nearest  face.     Locate  the  drilling  point, 
and  measure  the  length  of  the  wire  from  O  to  the  tower.     Scale  £  in.  = 
1ft. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES      61 

282.  From  F(2,  x-2)  in  the  plane  S(l  +  2)  3(3-2)  erect  a  perpendicular 
2  in.  long. 

283.  FromG(4  +  l,  x)  inT(4|  +  3)  2(4 \  —  2)  erect  a  perpendicular  l^in.  long. 

284.  From  K(3  +  l,  y)  in  W(f  +  !£)  2(5  —  3)  erect  a  perpendicular  2  in.  long. 

285.  From  M(3  +  i,  z)  in  X(  +  l$)co(  — 2)  erect  a  perpendicular  1$  in.  long. 

286.  From  N(2  +  l,  x)  in  R(l  +  3)  5(1 +  1£)  erect  a  perpendicular  2  in.  long. 

76.  EXERCISES  IN  SOLIDS 

In  the  following  exercises  only  one  or  two  solids  are  given  to  be  erected 
on  each  type  of  plane.  By  interchanging  solids  and  planes  in  different 
exercises,  a  large  number  of  exercises  may  be  very  easily  made  up.  Consider 
the  solids  opaque,  and  dot  all  invisible  lines. 

287.  Erect  an  equilateral  triangular  prism,   1-in.  sides  and  2-in.  altitude, 
on  the  plane  T(l  +2)  3(3-2),  with  the  center  of  the  base  at  A(2,  x-2) 
inT. 

288.  Erect  a  square  pyramid,  l|-in.  base  and  2-in.  altitude,  on  the  plane 
T(l  +  2)  3(3-3).     Center  of  base  at  B(2,  y-li). 

289.  Erect  a  1%-in.  cube  on  S(4£  +  3)  2(4£  - 2),  with  one  corner  at  C(4  +  i  y) 
and  another  corner  touching  H. 

290.  Erect  a  rectangular  pyramid,  base  1  in.   X    1£  in.,  altitude  2  in.,  on 
T(4£  +  3)  2(4$ -2),  one  corner  of  the  base  at  D(4  +  l,  x),  and  the  long 
edges  of  the  base  parallel  to  H. 

291.  Erect  a  pentagonal   pyramid,    base  radius   £   in.,  altitude  2  in.,  on 
R(£  +  H)  2(5-2),  center  of  base  at  E(3  +  l,  y). 

292.  Erect  a  hexagonal  prism,  base  l$-in.  long  diameter,  altitude  1$    in. 
on  R(f +  H)  2(5-3),  one  edge  of  the  base  parallel  to  V,  and  $  in. 
from  V. 

293.  Erect  a  hexagonal  pyramid,  1^-in.  short  diameter  of  base,  2$-in.  alti- 
tude, on  P(4  +  3)  4(4-3),  with  the  center  of  the  base  at  M(4-l  +  l|), 
and  one  edge  parallel  to  H. 

294.  Erect  a  l|-in.  cube  on  the  plane  Q(  +  l$)oo(  — 1$),  with  the  center  of 
the  base  at  N(3  +  l,  x),  and  the  diagonal  parallel  to  H. 

295.  Erect  a  cone,  l$-in.  base  and  2-in.  altitude,  on  the  plane  Q(  +  l$)oo 
(-1$),  with  the  center  of  the  base  at  N(3  +  l,  x). 

296.  Erect  a  cylinder,  1-in.  base  and  2-in.  altitude,  on  the  plane  X(l  +  3) 
5(1  +  1$),  with  the  center  of  the  base  at  0(2  +  1$,  x)  in  the  plane  X. 

297.  The    points  A(H  +  1-|),    B(lf  +  i-|)   and  C(2}  +  f—J)   limit  the 
base  of  a  triangular  prism  of  2-in.  altitude.     Draw  its  projections. 

298.  The  line  D(2  +  2  - 1)  E(l$  + 1  -  $)  is  one  edge  of  the  base  of  a  rectang- 
ular prism.     The  opposite  edge  runs  through  the  point  F(2  +  l  — $). 
Its  altitude  is  2  in.     Find  its  projections. 

299.  The  lines  A(3  + 1-1)  B(2  +  £-$)  and  BC(3  +  0-£)  are  two  sides  of  a 
parallelogram,  which  is  the  base  of  a  prism  2  in.  long.     Draw  the 
projections  of  the  prism. 

300.  The  apex  of  a  pyramid  is  A(2  +  2  — 1),  and  its  base  is  a  1  in.Xl$  in. 
rectangle  in  the  plane  T(4  +  3)  2(4  —  2).     Draw  its  projections. 

301.  The  apex  of  a  cone  of  l$-in.  base  diameter  is  M(3  +  2  —  2).     Its  base 
is  in  S(l+2)  3(3  —  3).     Draw  its  projections. 


62 


302. 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


303. 


0(2+1  —  1)  P(3$  4-  1  -  1)  is  the  edge  of  a  cube.     One  edge  of  the  cube 
is  in  H.     Draw  its  projections. 

The  apex  of  a  l^-in.  square  pyramid  is  A(3  +  2  —  2),  and  its  base  is  in 
R(  +  2)  oo(  —  1).     Draw  its  projections. 

304.  Draw   a   line   from   A(3  +  £-!£),    that   shall   intersect    M(14  +  |-2) 
—  1)  in  a   point   P,    and  be  1|   in.    distant   from   the   point 


77.  Problem  27.  —  To  pass  a  plane  through  a  given  point  per- 
pendicular to  a  given  line. 

This  problem  is  similar  to  Problem  24,  in  that,  the  required 
plane  being  perpendicular  to  the  given  line,  its  traces  must  be 
respectively  perpendicular  to  the  projections  of  the  given  line. 
Therefore,  we  know  the  direction  of  its  traces. 


FIG.  40. 


Analysis. — 1.  Draw  a  line  through  the  given  point  perpen- 
dicular to  the  given  line  and  parallel  to  H,  or  it  could  be  V. 
See  Theorem  XX,  Art,  24. 

2.  Find  the  F-piercing  point  of  this  auxiliary  line. 

3.  Pass  a  plane  through  the  auxiliary  line,  whose  traces  are 
respectively  perpendicular  to  the  projections  of  the  line. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES      63 

Proof. — It  can  be  proved  by  the  answers  to  these  two  ques- 
tions: Does  this  plane  contain  the  given  point?  Is  this  plane 
perpendicular  to  the  given  line? 

Construction. — Let  it  be  required  to  pass  a  plane  through  O 
(Fig.  40)  perpendicular  to  AB. 

1.  Draw  op  perpendicular  to  ab. 

2.  Draw  o'p'  parallel  to  GL. 

3.  Find  P,  the  ^-piercing  point  of  OP. 

4.  Through  p'  draw  t'T  perpendicular  to  a'b'. 

5.  Through  T  draw  Tt  perpendicular  to  ab.     The  plane  T 
will  contain  the  point  O,  and  will  be  perpendicular  to  AB. 

Notes. — 1.  If  the  given  line  lies  in  a  profile  plane,  the  problem  can  only 
be  solved  by  the  P-projection. 

2.  Under  certain  conditions  (that  is,  where  T  falls  outside  the  limits  of 
the  problem)  it  will  be  necessary  to  draw  two  lines  like  OP,  one  parallel  to 
V,  and  the  other  parallel  to  H. 

78.  Problem  28. — To  find  the  distance  from  a  point  to  a  plane. 

Analysis. — 1.  Erect  a  perpendicular  from  the  point  to  the 
plane. 

2.  Find    the    piercing    point  of  the  perpendicular  with  the 
plane. 

3.  Measure  the  distance  from  the  given  point  to  the  piercing 
point.     This  will  be  the  distance. 

Let  the  student  make  the  construction. 

79.  Problem  29. — To  project  a  line  on  any  oblique  plane. 

The  projection  of  any  point  on  any  plane  is  the  point  in 
which  that  plane  is  pierced  by  the  perpendicular  from  the  given 
point. 

Analysis. — 1.  Let  fall  perpendiculars  from  any  two  points 
(preferably  the  limiting  points)  of  the  line  to  the  plane. 

2.  Find  the  piercing  points  of  the  perpendiculars   with   the 
given  plane. 

3.  Join  these  piercing  points  with  a  line.     This  line  will  be  the 
required  projection. 

Let  the  student  make  the  construction. 

Check. — It  is  obvious  that  the  projection  of  a  point  lying  in  a 
plane  is  the  point  itself.  It  is  also  obvious  that  the  point,  in 
which  a  lin^  pierces  a  plane,  is  identical  with  the  projection  of 
that  point.  Therefore  the  line  will  intersect  its  projection, 
on  any  plane  at  its  piercing  point  with  that  plane.  Hence, 


64 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


it  is  probable  that  the  projection  is  correctly  made,  if  the  line 
and  its  projection  intersect 'at  the  piercing  point. 

80.  Problem  30. — To  measure  the  angle  between  any  line 
and  any  plane. 

First  Method. — The  measure  of  the  angle  between  a  line 
and  a  plane  is  the  angle  between  the  line  and  its  projection  on 
that  plane.  Therefore  one  analysis  suggests  itself;  as  follows, 

Analysis. — 1.  Project  the  line  on  the  plane  by  Problem  29. 

2.  Measure  the  angle  between  the  given  line  and  its  projec- 
tion thus  found,  by  Problem  14. 

Let  the  student  make  the  construction. 


b, 


t' 


81.  Second  Method. — This  method,  while  not  so  obvious  is 
much  quicker  and  better. 

Analysis. — 1.  From  any  point  in  the  given  line  erect  a  per- 
pendicular to  the  given  plane. 

2.  Measure  the  angle  between  the  given  line  and  this  perpen- 
dicular. 

3.  Subtract  this  angle  from   90°,   which   gives   the   required 
angle  between  the  line  and  plane. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       65 

Proof. — By  letting  fall  a  perpendicular  to  the  plane,  a  right- 
angled  triangle  is  formed  between  the  given  line,  its  projection, 
and  the  perpendicular.  The  angles  between  the  line  and  its 
projection,  and  between  the  line  and  the  perpendicular  are  com- 
plementary, and  add  up  to  90°.  Hence  this  method  might  be 
called  the  "  complementary  method."- 

Construction. — Let  it  be  required  to  measure  the  angle  between 
AB  (Fig.  41)  and  the  plane  T. 

1.  From  B  let  fall  a  perpendicular,  b'p',  bp,  to  T. 

2.  Find  the  //"-piercing  points  (or  F-piercing  points,  if  more 
convenient)  of  the  lines  AB  and  BP. 

3.  Revolve  B  into  H  about  the  #-trace  of  the  plane  Q  thus 
found. 

4.  Lay  off  a  perpendicular,  ^1x4,  to  pbt,  and  6  will  be  the 
required  angle. 

Compare  this  method  of  obtaining  the  angle  with  the  "  natural" 
method  on  the  same  exercise,  and  a  considerable  saving  will  be 
shown. 

82.  EXERCISES 

305.  Draw  the  traces  of  a  plane  T,  perpendicular  to  A(2  +  £  —  *)    B(4+-2 
-1),  through  the  point  0(1  +  1*- 2). 

306.  Draw  the  traces  of  a  plane  S,  perpendicular  to  0(2  +  3-1)  D(4  +  l  — 3), 
through  the  point  P(3  + 1  -  3). 

307.  Draw  the  traces  of  a  plane  X,  perpendicular  to  E(2  +  1  —  1)  F(4  +  1  —  2), 
through  the  point  M(3  +  l-3). 

308.  Draw  the  traces  of  a  plane  W,  perpendicular  to  G(3  +  *  — 1)  K(3  +  2  — 2), 
through  the  point  P(3  +  3-l). 

309.  Draw  the  traces  of  a  plane  Y,  perpendicular  to  A(2  + 1  —  1)  B(4  + 1*  —  2) 
at  its  middle  point. 

310.  Draw  the  traces  of  a  plane  Q,  perpendicular  to  0(2  +  1-3)  D(4  +  2-l), 
through  the  point  D. 

311.  E(2  +  3-3)  F(4  +  l  — 2)   is  the  base  of  an  isosceles  triangle,  whose 
vertex  is  in  GL.     Draw  its  projections  and  revolve  to  show  its  true 
size. 

312.  G(2*  + 1  —  2)    K(3*  +  *  —  1)  is  one  edge  of  a  cube,  one  of  whose  corners 
rests  on  H.     Draw  the  projections  of  the  cube.     Dot  all  invisible  lines. 

313.  The  center  line  of  a  stick  1*  in.  square  is  the  line  A(2  +  l  -  l)B(4  +  3  — 3). 
Two    i-in.   holes  are   drilled   straight  through   the   stick,    f  in.  from 
the  respective  ends.     Draw  the  projections  of  the  stick  and  holes, 
dotting  all  invisible  lines. 

314.  Draw  the  projections  of  a  hexagonal  prism,  1^-in.  long  diameter,  1*- 
in.  altitude,  whose  center  line  is  0(2  +  3-1)  D(4  +  l— 3). 

315.  E(3  +  2-i)  F(3  +  i-2^)  is  the  axis  of  a  2-in.  cylinder.     Draw  the 
projections  of  the  cylinder. 

5 


66  PRACTICAL  DESCRIPTIVE  GEOMETRY 

316.  G(2  +  $-l)  K(4  +  2-2)  is  the  axis  of  a  right  circular  cone  of  2-in. 
base.     Draw  its  projections. 

317.  The  upper  surface  of  a  vein  of  coal  is  found  to  be  perpendicular  to  a 
shaft  bored  in  the  direction  of  A(2$  +  1-1)  B(4  +  3  -  2).     The  surface 
is  located  at  C(3,  x,  y)  on  AB.     Find  the  traces  of  the  plane  of  the  ore. 

318.  A  guy-wire  supporting  a  stack  runs  from   D(lJ  +  3  —  1)  on  the  stack  to 
E(4  +  l  — 2)  on  a  roof  perpendicular  to  the  wire.     Find  the  traces  of 
the  plane  of  the  roof. 

319.  A  shaft  center  line  isG(2$  +  l-3)  K(2$  +  3-l$).     A   12-in.    pulley, 
with  6-in.  face,  is  located  in  the  center  of  GK.     Draw  the  projections  of 
the  pulley  (solid,  without  spokes),  to  a  scale  of  2  in.  =  1  ft. 

320.  Find  the  distance  from  A(4$  +  2-1)  to  the  plane  T(l +  3)  4$(4$-3). 

321.  Find  the  distance  from  B(4  +  2-l)  to  the  plane  8(1  +  3)  4(1-3).     ' 

322.  Find  the  distance  from  C(4  +  2-2)  to  the  plane  R(2  +  3)  3(2  —  2$). 

323.  Find  the  distance  from  D(3  +  3-3)  to  the  plane  Q(  +  2)oo(-l). 

324.  Find  the  distance  from  E(5  +  l-l)  to  the  plane  W(l+2)  3(5-2). 

325.  Find  the  distance  from  F  (3$ +  !$-£)  to  the  plane  X(2  +  3)  5(2-$). 

326.  Find  the  distance  from  G(3  +  l-l)  to  the  plane  Y(l +3)(5  +  2)(l -3) 
(5-2). 

327.  A  3-in.  cube  is  placed  on  H,  with  one  face'against  V.     Find  the  distance 
from  the  upper  rear  right-hand  corner  to  the  plane  of  the  three  adjacent 
corners. 

328.  An  inclined  roof  is  in  the  plane  T(l+2)    5(1  —  3).     From  a  point 

M(2,  0—1)  in  the  floor  it  desired  to  run  a  cylindrical  fine  8-in. 
diameter. through  the  roof,  at  right  angles  to  it.  What  is  the  length 
of  the  flue?  Draw  the  projections  of  the  opening  in  the  roof.  Scale 
1$  in.  =  1  ft. 

329.  Project  M(2  +  2-l)  N(4|  +  2-l)  on  the  plane  T(5  +  3)  1(5-3),  and 
find  the  true  length  of  the  projection. 

330.  Project  A(2f  +  $-U)   B(3$  +  l$-i)  on  S(4  +  l|)  3(2-H),  and  find 
where  AB  pierces  S. 

331.  Project  C(3  +  l-$)  D(4  +  2-2)  on  R(  + l)oo(-l$). 

332.  Project  0(3  +  1-$)  D(4  +  2-2)  on  P(4$  +  3)  4$(4$-3). 

333.  Project  E(3  +  2-2)  F(3  +  $-l)  on  Q(  +  l$)oo(-2). 

334.  Project  the  triangle  G(2  +  l-2)   H(3  +  2-l)   K(3$  +  !$-!$)   on  the 

plane  S  (  +  $)oo(-2$). 

335.  Measure    the   angle  that   A    (1  +  1|-1)     B(3 +  !$-!)    makes   with 
Q(l+3)  5(1-2). 

336.  Measure  the  angle  that  0(1+2-2)  D(3  +  l$-$)  makes  with  R(4  +  3) 
1(4-1). 

337.  Measure  the  angle  that  E(l+2-2)  F(4  +  l-l)  makes  with  S(3  +  3) 
3(5-2). 

338.  Measure  the  angle  that  G(l  +  3  -  2)  K(3  + 1  —  1)  makes  with  the  profile 
plane  at  P(3,  0,  0). 

339.  Measure  the  angle  that  M(2  +  2-3)  N(4  +  l-$)  makes  with  T(+l$) 
oo(-l). 

340.  Measure  the  angle  O(2  +  3-2)  P(3$  +  l-$)   makes  with  W(3$  +  3) 
3(3$-3). 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES      67 

341.  Measure  the  angle  that  A(2  +  l  —  $)  B(4  +  l  —  $)  makes  with  X(l  +  li) 
8i(5-l». 

342.  Measure  the  angle  that  0(3  +  1-1)  D(4  +  2-3)  makes  with  Y(l  +  3) 


343.  Measure  the  angles  that  E(2  +  2-2)  F(3£  +  J-1)  makes  with    H,  V 
and  P. 

344.  Measure  the  angle  that  the  center  line  of  the  flue  in  Exercise  328  makes 
with  the  floor. 

345.  Find  the  angles  that  the  guy-  wire  in  Exercise  318  makes  with  H  and  V. 

346.  In  Exercise  316,  what  angle  does  any  element  of  the  cone  make  with 
the  plane  of  the  base? 

347.  What  angles  does  the  telephone  wire  in  Exercise  281  make  with  H,  V 
andP? 

348.  What  angles  does  the  hip  rafter  EG  in  Fig.  25  make  with  the  roof  C, 
and  with  the  floor? 


83.  Problem  31.— To  draw  a  line  through  a  point  making  a 
given  angle  with  a  given  plane,  and  intersecting  a  given  line 
in  the  plane. 

This  problem  is  a  special  case  of  Problem  30,  and  would  have 
an  infinite  number  of  solutions,  but  for  the  last  condition  intro- 
duced; that  is,  intersecting  some  line  in  the  plane.  This  limits 
the  solution  to  two  possible  lines,  or  even  one,  and,  if  the  line 
be  too  far  away,  the  problem  is  impossible. 

Analysis. — The  locus  of  all  the  lines  making  the  given  angle 
with  the  given  plane  would  be  the  surface  of  a  right  circular 
cone,  its  apex  at  the  given  point,  and  its  base  in  the  given 
plane;  hence, 

1.  Let  fall  a  perpendicular  from  the  given  point  to  the  given 
plane,  and  find  its  piercing  point. 

2.  This  point  is  the  center  of  the  base  of  an  imaginary  cone; 
therefore  obtain  the  true  length  of  the  perpendicular. 

3.  With  the  revolved  perpendicular  as  one  side,  construct  a 
right-angled  triangle  with  the  given  angle  at  the  base. 

4.  With  the  base  of  this  triangle  as  a  radius,  draw  an  arc 
from  the  piercing  point  of  the  perpendicular  as  a  center,  inter- 
secting the  given  line. 

">.  Connect  the  given  point  with  the  intersection  of  the  arc 
and  the  given  line.  This  line  will  fulfil  the  conditions  of  the 
problem. 

Construction. — Let  it  be  required  to  draw  a  line  from  O  (Fig. 
42),  making  60°  with  T,  and  intersecting  the  line  AB,  that  lies 
inT. 


68 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


1.  Draw  OP,  perpendicular  to  T,  from  O. 

2.  Find  the  piercing  point,  P,  of  OP  with  T. 

3.  Measure  OP  (o^p'). 

4.  Construct  a  right  triangle  p'mo\,  with  the  angle  at  m  =  60°. 

5.  Revolve  P  and  AB  into  H  about  Tt. 

.   6.  From  p1  as  a  center  and  with  o^m  as  a  radius,   strike  an 
arc,  intersecting  bax  at  CA. 

7.  Counter-revolve  ct  to  cc'  on  AB. 

8.  Join  O  with  C.     OC  is  the  required  line. 


tXb 


FIG.  42. 


84.  Problem  32.— Draw  a  line  through  a  given  point  that  shall 
make  given  angles  with  H  and  V. 

Note. — The  sum  of  the  two  given  angles  may  be  any  amount  between  0° 
and  90°.  Any  line  parallel  to  GL  makes  angles  with  H  and  V  equal  to  zero, 
and  any  line  in  a  profile  plane  makes  angles  with  H  and  V  that  add  up  to  90°. 

Analysis. — The  analysis  of  this  problem,  and  several  problems 
later,  depends  on  certain  properties  of  the  right  circular  cone, 
so  it  is  best  to  discuss  these  before  undertaking  the  construction 
of  the  problem. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES      69 

DISCUSSION  OF  THE  RIGHT  CIRCULAR  CONE. 

Three  properties  of  the  cone  are  particularly  useful  in  the 
solution  of  certain  problems. 

.  1.  The  elements  of  the  cone  all  make  the  same  angle  with 
the  plane  of  the  base,  and  they  are  all  of  the  same  length, 
because  one  way  of  generating  a  cone  is  to  revolve  a  right- 
angled  triangle  about  one  of  its  perpendicular  sides. 

2.  All  planes  tangent  to  a  cone  make  the  same  angle  with  the 
plane  of  the  base,  because  a  tangent  plane  is  perpendicular  to 
the  generating  triangle  of  the  cone,  and  contains  its  hypothenuse. 

3.  Every  element  of  the  cone  and  every  plane  tangent  to  the 
cone  pass  through  the  apex. 

85.  Construction  of  Problem  32. — Let  it  be  required  to  draw  a 
line  through  A,  Fig.  43,  that  shall  make  45°  with  H,  and  30° 
with  V. 


FIG.  43. 

ex  of  a  right  circular  cone,  whose  base  is  in 
nts  make  45°  with  H. 

make  45°  with  H ',  and  all  pass  through  A. 
is  to  be  found  which  are  the  four  possible 
30°  with  V. 

a  line  AB  parallel  to  H,  making  30°  with  7, 
to  the  elements  of  the  cone,  its  projections 


70  PRACTICAL  DESCRIPTIVE  GEOMETRY 

3.  Revolve  B  about  an  axis  through  A  perpendicular  to  V, 
until  it  comes  into  contact  with  the  cone  at  abj  a'b'.  That  is, 
when  the  circular  path  of  B  cuts  the  circular  base  of  the  cone. 
AB  is  the  required  line. 


Notes.  —  1.  ABB!  may  be  regarded  as  part  of  the  surface  of  a  second  cone, 
and  AB  will  then  be  the  line  of  intersection  of  the  two  cones,  the  common 
element. 

2.  Four  lines  may  be  drawn  through  A,  all  making  45°  with  H,  and  30° 
with  V,  AB,  AC,  AD  and  AE,  all  shown  in  Fig.  43. 

86.  "*»     EXERCISES 

349.  Draw  a  line  from  A(4  +  2-l)  making  60°  with  T(l+3)  5(1-2),  and 
intersecting  B(2,  0,  x)  C(3,  y,  0)  in  T.     What  angle  does  the  line  make 
with  BC? 

350.  Draw  a  line  from  D  (3£  +  2  -2)   making  30°  with  S(l+3)   3(5-1), 
intersecting   E(2,    x,  0)  F(4J,  0,  y)  in  S.     What  angle  does  the  line 
make  with  EF? 

351.  Draw   a    line   from    G(3  +  2-l)    making   30°    with    R(  +  l|)co(-l), 
intersecting  H(2  +  l£,  0)  K(4,  0  —  1).     What  angle  does  the  line  make 
with  HK? 

352.  Through  A(3  +  2-l)  draw  four  lines,  each  making  angles  of  6010  with 
V  and  15°  with  H.     What  angles  do  they  make  with  one  another? 

353.  Through  B(3  +  !$-!£)  draw  four  lines,  each  making  angles  of  20° 
with  V  and  35°  with  H. 

354.  Draw  a  line  3|  in.  long,  terminating  in  H  and  V,  making  angles  of  45° 
with  V,  and  30°  with  H. 

355.  Draw  a  line  through  0(4+1-1)  making  50°  with  H  and  40°  with  V. 

356.  Draw  a  line  through  D(3,  0,  0)  making  angles  of  45°  with  H,  and  30° 
with  V. 

357.  Through   E(l  +  l|-l)    F(4  +  1J-1)    pass   a   plane   perpendicular   to 
T(l+3)  5(1-2). 

358.  Through    G(2  +  3-3)    H(4  +  1-J)    pass    a    plane    perpendicular    to 
S(  +  2)oo(-l). 

359.  From  a  point  K(16  ft.  +24  ft.  —  16  ft.)  on  a  stack,  a  guy-  wire  is  run  to 
a  point  on  a  horizontal  roof  8  ft.  above  H.     The   guy-wire   inclines 
60°  to  H  and  20°  to   V,  running  to  the  southeast.     Where  does  it 
strike  the  roof,  and  what  is  its  length?     Scale  \  in.  =  1  ft. 

87.  Problem  33.  —  To  pass  a  plane  parallel  to  a  given  plane 
through  a  given  point. 

Two  parallel  planes  must  have  their  respective  traces  parallel. 
The  converse  is  not  always  true.  For  example,  witm  ss  any  two 
planes,  whose  traces  are  parallel  to  GL.  They  may,  or  may  not, 
be  parallel. 

Analysis.  —  1.  Draw  a  line  through  the  given  point  parallel  to 
the  J^-trace  of  the  given  plane. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       71 

2.  Through  the  F-piercing  point    of  this  auxiliary  line  draw 
the  7-trace  of  the  plane  parallel  to  the  V-trace  of  the  given  plane. 

3.  From  the  intersection  of  the  V-trace  and  GL  draw  the 
//"-trace  parallel  to  the  #-trace  of  the  given  plane. 

Construction. — Let  the  student  make  the  construction  and 
draw  the  figure.  Note  the  similarity  in  this  problem  and  in 
Problem  27. 

Note. — If  the  GL  intersection  of  the  traces  is  not  accessible,  two  auxiliary 
lines  may  have  to  be  drawn. 

Special  Case. — When  the  given  plane  is  parallel  to  GL.     In 

this  case  the  profile  view  must  be  used.     The  P-traces  of  the 
given  and  required  planes  must  be  parallel. 

88.  Problem  34. — To  pass  a  plane  parallel  to  a  given  plane  at 
a  given  distance  from  it. 

Analysis. — 1.  From  any  point  in  the  given  plane  erect  a  per- 
pendicular of  a  length  equal  to  the  given  distance  according 
to  Problem  25. 

2.  At  the  other  end  of  this  perpendicular  draw  a  plane  parallel 
to  the  given  plane,  according  to  Problem  33. 

The  similarity  between  this  problem  and  the  preceding  one  is 
so  obvious  that  no  construction  or  figure  is  necessary. 

Second  Method.  Analysis. — 1.  Pass  an  auxiliary  plane  per- 
pendicular to  the  #-trace  of  the  given  plane. 

2.  Revolve   the   intersection   of   the   two   planes    about   the 
H-tr&ce  of  the  auxiliary  plane. 

3.  Draw  a  parallel  to  the  revolved  intersection  at  the  required 
distance. 

4.  Counter-revolve   this   line    and   draw   the  required  traces 
through  its  piercing  points,  parallel  to  the  given  traces. 

This  method  will  be  found  easy  and  convenient.  Let  the 
student  make  the  construction. 

89.  EXERCISES 

360.  Through  A(4  +  1  -  1)  pass  a  plane  parallel  to  T(l  +3)  3(1  -  1).     What 
is  the  distance  between  the  planes? 

361.  Through    B(2  +  l-2)    pass    a    plane    parallel    to    S<1  +  1)    5i(l+3). 
What  is  the  distance  between  the  planes? 

362.  Through  0(5  +  1-2)  pass  a  plane  parallel  to  R(2  +  3)  2(4-3).     What 
is  the  distance  between  the  planes? 

363.  Through  D(5  +  l-l)  pass  a  plane  parallel  to  Q(4  +  3)  3(2-3).     What 
is  the  distance  between  the  planes? 


72  PRACTICAL  DESCRIPTIVE  GEOMETRY 

364.  Through    E(4 +  !£-£)    pass    a    plane    parallel    to    X(l+3)    4(1-2). 
What  is  the  distance  between  the  planes? 

365.  Through  F(5  +  l-2)  pass  a  plane  parallel  to  W(  +  2)oo(-l).     What 
is  the  distance  between  the  planes? 

366.  Draw  the  traces  of  a  plane  1  in.  from  Q(l  +3)  3(1  —  1)  and  parallel  to  it. 

367.  Draw  the  traces  of  a  plane  1£  in.  from  R(2  +  3)  2(4  —  2)  and  parallel 
to  it. 

368.  Draw  the  traces  of  a  plane  1  in.  from  S(l  +  l)  5$(l+3)  and  parallel 
to  it. 

369.  Draw  the  traces  of  a  plane  l{  in.  from  T(4  +  2)  3(2-2)  and  parallel  to 
it. 

370.  Draw  the  traces  of  a  plane  1  in.  from  W(l  +  1)5(1  —  3)  and  parallel  to  it. 

371.  A  saw-toothed  roof  is  being  erected  for  a  greenhouse.     The  plane  of 
one  roof  plane  is  T(l  +2)  3(1  —2J).     Find  the  traces  of  two  other  roof 
planes,  the  distance  between  any  two  adjacent  parallel  planes  being 
7  ft.     Draw  to  a  scale  of  li  in.  =  1  ft. 

372.  The  alternate  planes  in  the  roof,  in  Exercise  371  are  perpendicular  to 
those  already  found.     The  ridge  line  is  a  horizontal  line,  8  ft.  above 
H.     Construct   the  H  and  F-projections  of  as  much  of  the  roof  as 
possible  in  the  space.     Scale  |  in.  ==  1  ft. 

90.  Problem  35. — To  find  the  true  angle  between  two  planes. 

The  angle  between  two  planes  is  considered  to  be  equal  to 
that  angle  included  between  the  perpendiculars  to  the  line  of 
intersection,  that  lie  in  each  plane.  It  is  evident,  then,  that 
the  plane  of  the  measuring  angle  is  perpendicular  to  the  line 
of  intersection,  and  therefore  to  both  of  the  given  planes.  Two 
methods  of  finding  this  angle  are  given  herewith. 

First  Method.  Analysis. — 1.  Find  the  intersection  of  the 
given  planes. 

2.'  At  any  assumed  point  on  the  line  of  intersection  pass  a 
plane  perpendicular  to  it. 

Note.— Only  one  trace  of  this  auxiliary  plane  is  usually  necessary. 

3.  Measure  the  angle  between  the  intersections  of  the  auxiliary 
plane  with  the  two  given  planes. 

Construction. — -Let  S  and  T  (Fig.  44),  be  the  two  planes. 

1.  Obtain  their  line  of  intersection,  AB. 

2.  Through  O,  assumed  on  AB,  draw  OP,  perpendicular  to 
AB  and  parallel  to  V. 

3.  Obtain  P,  the  //-piercing  point  of  OP. 

4.  Draw  qq  through  p,  perpendicular  to  ab. 

5.  Join  o  with  x  and  y  (where  qq  crosses  Tt  and  Ss,  respec- 
tively).    OX  and  OY  are  the  perpendiculars  to  AB,  lying  in 
T  and  S. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       73 

6.  Revolve  O  about  qq  into  H,  and  connect  ot  with  x  and  y. 
The  angle  0  (xoxy)    is  the  true  measure  of  the  angle  between 
the  planes. 


91.     Second  Method.     Analysis. — 1.    From  any  point  let  fall 
two  perpendiculars,  one  to  each  of  the  two  given  planes. 

2.  Measure  the  angle  included  between  these  perpendiculars. 
This  will  be  the  correct  measure  of  the  required  dihedral  angle. 


•S 


FIG.  4.ri. 

Proof. — The  two  perpendiculars  from  the  point  to  the  planes 
determine  a  plane  perpendicular  to  both  planes.  Fig.  45  shows 
a  view  of  two  intersecting  planes  S  and  T,  looking  along  their 
intersection.  O  is  taken  anywhere,  and  perpendiculars  OM  and 
ON  are  let  fall  to  T  and  S  respectively.  The  plane  of  MON  cuts 


74 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


lines  MP  and  NP  from  T  and  S,  and  the  four  lines,  OM,  ON, 
MP  and  NP  form  a  quadrilateral,  the  sum  of  whose  angles  is 
360°.  Prove  this.  The  angles  at  M  and  N  are  right  angles 
by  construction,  and  therefore  add  up  to  180°.  This  leaves 
180°  as  the  sum  of  angle's  </>  and  6,  at  O  and  P  respectively. 
The  angle  </>  is  therefore  the  supplement  of  6,  which  is  the 
angle  between  T  and  S.  But  the  measure  of  any  angle,  plane 
or  dihedral,  may  be  either  the  acute  angle  or  its  supplement, 
so  the  angle  between  the  planes  S  and  T  may  be  either  6  and  0, 
and  as  </>  is  the  angle  between  the  perpendiculars,  it  is  also  the 
measure  of  the  given  dihedral  angle. 

Note. — Although  the  angle  between  two  lines,  or  two  planes,  may  be 
either  the  acute  or  the  obtuse  angle,  as  stated  in/  the  foregoing,  the  acute 
angle  is  usually  taken  as  the  true  measure. 


FIG.  46. 

92.  Problem  36. — To  find  the  true  angle  between  a  plane  and 
one  of  the  projecting  planes. 

Analysis. — To  find  the  angle  between  T  and  H. 

1.  Let  any  point  in  the  F-trace  of  T  be  the  apex  of  a  right 
circular  cone,  whose  base  is  in  H. 

2.  Let  the  circular  base  of  the  cone  be  tangent  to  the  //-trace  of  T. 

3.  The  angle  of  the  elements  with  H  will  be  the  angle  between 
T  and  H. 

Referring  to  Art.  84,  the  Discussion  of  the  Cone,  we  shall  find 
that  any  plane  tangent  to  a  cone  makes  the  same  angle  with  the 
plane  of  the  base  that  is  made  by  the  elements  with  the  base.  . 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       75 

Fig.  46  shows  such  a  cone  with  its  apex  at  o'  in  the  F-trace, 
and  its  base  tangent  to  the  //-trace.  The  angle  6  is  therefore 
the  true  angle  between  T  and  H. 

Fig.  46  also  shows  a  cone  with  its  apex  at  m  in  the  //-trace  of 
T,  and  its  base  tangent  to  the  F-trace.  Its  angle,  (j>,  is  the  true 
angle  between  T  and  V. 

93.  Problem  37. — Given  one  trace  of  a  plane  and  its  angle 
with  the  plane  of  projection,  to  find  the  other  trace. 

Analysis. — 1.  Construct  a  cone,  whose  elements  make  the  given 
angle  with  its  base,  tangent  to  the  given  trace,  with  the  center 
of  its  base  in  GL. 

2.  Draw  the  other  trace  through  the  apex  of  the  cone  and  the 
intersection  of  the  given  trace  with  GL. 

Notes. — 1.  If  the  given  trace  does  not  intersect  GL  within  bounds,  two 
cones  can  be  constructed,  and  the  trace  drawn  through  the  two  vertices. 

2.  If  the  given  trace  is  parallel  to  GL,  the  other  trace  will  also  be  parallel, 
and  the  quickest  solution  is  effected  through  the  profile  plane. 

3.  By  inverting  the  cone,  another  plane  could  be  found  that  would  fulfil 
the  conditions. 

94.  Problem  38. — The  H-trace  of  a  plane  being  given,  and  its 
angle  with  V,  to  find  the  V-trace. 

Analysis. — 1.  From  any  point  in  the  given  //-trace  draw  a 
line  making  the  given  angle  with  GL. 

2.  From  the  same  point  in  the  //-trace,  draw  a  line  perpen- 
dicular to  GL. 

3.  Draw  a  semi-circle  in  V,  having  the  second  point  on  'GL 
as  its  center,  and  the  first  point  on  GL  as  the  other  end  of  the 
radius. 

4.  Draw  the   F-trace  tangent  to  the  circle  from  the  inter- 
section of  the  //-trace  and  GL. 

Note. — This  solution  is  the  same  cone  construction  as  employed*  in 
previous  problems. 

Construction.— Let  Tt,  Fig.  46,  be  the  given  trace,  and  let 
(f>  be  the  angle  the  plane  makes  with  V. 

1.  Assume  M,  any  point  on  Tt. 

2.  Lay   off  mn,  making  the  angle  0  with  GL. 

3.  With  m'  as  a  center  and  m'n'  as  a  radius,  draw  the  semi- 
circle, the  base  of  the  cone. 

4.  From  T  draw  a  tangent  to  the  semi-circle. 
This  will  be  the  required  F-trace. 


76  PRACTICAL  DESCRIPTIVE  GEOMETRY 

Note. — In  this  problem,  as  in  Problem  37,  two  solutions  are  always  pos- 
sible, except  when  the  angle  given  is  a  right  angle.  This  is  accomplished  by 
inverting  the  cone  in  Problem  37;  that  is,  using  the  same  base,  but  drawing 
the  apex  below  H;  and  in  Problem  38  it  is  effected  by  completing  the  circle, 
and  drawing  the  tangent  on  the  opposite  side. 

95.  EXERCISES 

Find  the  true  dihedral  angle  between  the  planes  given  in  the  following 
exercises. 

373.  T(4  +  3)  2(5-2)  and  S(l+3)  5(1-2.}). 

374.  Q(l+2$)  4$(l-2)  and  R(4  +  3)  2(2-3). 

375.  W(  +  l$)co(-l)  and  X(5'+3)  2(5-1$). 

376.  Y(  +  l$)oo(-l)  and  Z(l+2$)  3(5-2}).' 

377.  Q(2  +  3)  4(5-1$)  and  R(5  +  3)  2f($-3). 

378.  S(  +  2)oo(-$)andT(  +  f)oo(-lf). 

379.  W(3  +  3)  3(3-3)  and  X(l  +  2)  5(1  +  1). 

380.  Y(l+3)  3(1-2)  and  Z(5  +  2)  3(5-1$). 

381.  Q(l+3)  2$(1-1)  and  R(4  +  3)  3(5-2). 

382.  S(  +  2)oo(-l)  andP(3  +  3)  3(3-3),  H  and  V. 

383.  T(4  +  3)  2(5-2)  and  H,  V  and  P. 

384.  Q(  +  l$)oo(-2)and#andF. 

385.  S(l  +  2)  5(1  +  1)  and  H,  V  and  P. 

386.  W(l+2)  4(1-3)  and  X(5  +  3)  4(1-3). 

387.  Given  A(2,  0-1$)  B(3  +  l,  0)  and  a  plane  S(3  +  l)  5(2-1$).     Pass  a 
plane  T  through  AB  that  will  make  60°  with  S. 

388.  Pass  a  plane  Q  through  C(2  +  2-$)  D(3  +  $-l$)  which  will  make  45° 
with  the  plane  R,  which  contains  CD  and  is  parallel  to  GL.     Find  the 
traces  of  Q.. 

389.  Find  the  traces  of  two  planes  containing  the  line  C(2  + 2  —  $)  D(3-f$ 
—  1$),   each  making  60°   with   H.     Measure  the  angle  between  these 
planes. 

390.  Through  the  line  E(3  +  $  -  2)  D(4$  +  If  -  $)  pass  a  plane  S  making  60° 
with  H,  and  a  plane  T  that  makes  60°  with  S. 

391.  Measure  all  the  dihedral  angles  of  the  tetrahedron  A(2,  0  - 1)  B  (4,  0  - 1) 
C(4,  0-2i)  D(2$  +  l$-2). 

392.  Measure  the  hip  roof  angle  between  the  roofs  C  and  D  in  Fig.  25, 
Article  54. 

393.  A  square  chute  has  its   center  line  A(4  +  l$-$)    B(2,   0-2).     The 
diagonal  of  its  right  section  is  parallel  to  H,  and  its  sides  are  each  12  in. 
in  width.     What  angle  does  any  one  of  its  sides  make  with  the  floor? 
Draw  the  intersection  of  the  chute  with  H.     Draw  the  chute  to  a 
scale  of  1$  in,  =1  ft. 

394.  A  tower  is  designed  in  the  form  of  a  hexagonal  pyramid,  having  a 
16  ft.  base  and  an  altitude  of  20  ft.     The  framework  is  of  structural 
iron.     At  what  angles  must  the  angle  irons  for  base  and  rafters  be 
rolled? 

395.  Draw  the  traces  of  two  planes,  S  and  T,  making  angles  of  45°  with 
each  other,  containing  the  line  M(2,.0-2)  N(3  +  l$,  0). 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       77 

Note.  —  Projecting  planes  are  not  intended  to  be  used  as  either  S  or  T  in 
this  exercise. 

396.  Find  the  traces  of  two  planes  making  60°  with  each  other  each  contain- 
ing the  line  O(2  +  l  —  1)  P(4  +  l  —  1),  one  of  them  being  parallel  to 


397.  A  plane  makes  30°  with  H,  and  its  #-trace  is  T(4,  0,  0)  t'(l,  0-2). 
Find  its  F-trace. 

398.  Two  planes,  Q  and  R,  have  their  ^-traces  in  the  line  0(4,  0,  0)  D(2,  0 
-2).     Q  makes  45°  with  H,  and  R  makes  75°  with  H.     Draw  their 
F-traces  and  measure  the  angle  between  Q  and  R. 

399.  Draw  the  traces  of  the  bisecting  plane  between  Q  and  R  in  Exercise 
398. 

400.  The  H-tT&ce  of  a  plane  T  is  A(2,  0  -  J)  B(4,  0  -  £)  and  the  plane  makes 
15°  with  V.     Find  the  F-trace.     Another  such  plane  could  be  found. 
Draw  its  F-trace. 

401.  The  #-trace  of  two  planes,  S  and  R,  is  0(5,  0,  0)  D(2,  0-2).     Both 
planes  make  60°  with  H.     Draw  their  F-traces,  measure  the  angle 
between  them,  and  draw  the  F-trace  of  the  bisecting  plane  of  the 
angle. 

402.  The  F-trace  of  two  planes,  X  and  Y  is  E(4,  0,  0)  F(2  +  li,  0).     Both 
planes  make  45°  with  H.     Draw  their  £T-traces. 

403.  The  F-trace  of  two  planes,  Q  and  R,  is  the  line  G(2-t-3,  0)  K(4  +  2,  0). 
Both  planes  make  30°  with  F.     Draw  their  ^-traces.     Measure  the 
angle  between  them. 

404.  A  horizontal  line  in  a  vein  of  ore,  that  inclines  15°  to  H,  runs  from 
A  (4,  0,  0)  south,  42°  west.     Find  the  F-trace  of  the  plane  of  the  ore. 

405.  A(2,  0-2)  B(5,  0,  0)  is  the  line  of  the  lower  edge  of  a  roof  inclined  30° 
from  the  horizontal.     The  length  of  the  roof  is  30  ft.,  with  a  corner  at 
A,    and    the    ridge    is    10   ft.    above   the   lower   edge.     The   roof  is 
rectangular.     Draw  its  H-  and  F-projections  to  a  convenient  scale. 

96.  Problem  39.  —  To  find  the  traces  of  a  plane  passed  through 
a  given  line  in  a  given  plane,  so  as  to  make  a  given  angle  with 
the  given  plane. 

Analysis.  —  Assume  a  point  in  the  given  line,  and  pass  a  plane 
through  the  point  perpendicular  to  the  line. 

2.  Revolve  the  line  cut  by  this  auxiliary  plane  from  the  given 
plane  about  the  trace  of  "  the  auxiliary  plane  into  H  or  V,  as  the 
case  may  require. 

3.  From  the  revolved  position  of  the  assumed  point  on  the 
given  line  lay  off  a  line,  making  the  given  angle  with  the  revolved 
line. 

4.  Where  this  new  line   crosses  the  trace  of    the   auxiliary 
plane  will  be  a  point  on  the  trace  of  the  required  plane. 

5.  Since  the  required  plane  must  contain  the  given  line,  its 
traces  will  pass  through  the  H-  and   F-piercing  points  of  the 


78  PRACTICAL  DESCRIPTIVE  GEOMETRY 

given  line.  As  we  have  these  two  points,  and  also  the  point  on 
the  required  trace  determined  in  operation  (4),  we  have  the  three 
points  necessary  to  determine  the  traces  of  the  required  plane. 

Construction.— Refer  to  Fig.  44,  Article  90.  Let  S  be  the 
given  plane,  AB  the  given  line  in  S,  and  6  the  given  angle. 

1.  Find  the  piercing  points  of  AB. 

2.  Assume  O  on  AB. 

3.  Draw  the  //-trace  of    Q,   containing  O,    perpendicular  to 
AB. 

4.  Revolve  OY  about  qq  to  oLy  in  H. 

5.  Lay  off  from  ol  a  line  otx,  which  makes  6  with  o^. 

6.  Draw  the  //-trace  of  T  through  b  and  x.  , 

7.  Draw  the  F-trace  of  T  through  T  and  a'. 

Notes. — 1.  Two  planes,  such  as  T,  making  0  with  S  can  always  be  drawn, 
except  when  6  is  90°. 

2.  When  it  is  required  to  pass  such  a  plane  perpendicular  to  the  given 
plane,  it  is  only  necessary  to  draw  a  line  perpendicular  to  the  given  plane  at 
any  point  of  the  given  line.  The  given  line  and  its  perpendicular  intersector 
then  determine  the  required  plane. 

97.  Problem  40. — To  find  the  traces  of  a  plane  passed  through 
a  given  line,  so  as  to  make  any  required  angle  with  H  or  V. 

Limitation. — The  given  angle  must  be  as  large  or  larger  than 
the  angle  that  the  given  line  makes  with  the  plane. 

Analysis. — 1.  Take  any  point  on  the  line  for  the  apex  of  a  cone, 
having  its  base  in  the  plane  to  which  the  angle  is  given,  its  ele- 
ments making  the  given  angle  with  that  plane.  See  Article  84. 

2.  Find  the  piercing  points  of  the  given  line  and  draw  the 
traces  of  the  plane  which  contains  the  line  and  is  tangent  to  the 
cone. 

Let  the  student  make  the  construction. 

98.  Problem  41. — To  find  the  traces  of  a  plane  which  shall 
make  any  required  angles  with  H  and  V. 

Limitations.-— The  sum  of  these  angles  must  be  equal  to  or 
more  than  90°,  or  equal  to  or  less  than  180°.  A  plane  parallel 
to  H  is  90°  from  V  and  0°  from  //.  A  profile  plane  is  90°  from 
V  and  90°  from  //.  Oblique  planes  will  be  between  these 
limits. 

Construction. — Let  it  be  required  to  draw  a  plane  making 
60°  with  H  and  45°  with  V. 

1.  Draw  the  projections  of  a  sphere,  having  any  radius,  and 
its  center  at  any  point  in  GL.  (See  Fig.  47.) 


.      PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       79< 

2.  Draw  a  60°  cone  tangent  to  the  sphere  with  its  base  in  H 
and  its  apex  in  V  at  a'. 

3.  Draw  a  45°  cone  tangent  to  the  sphere  with  its  base  in  V 
and  its  apex  in  H  at  b. 

4.  Draw  the  F-trace  of  the  required  plane  through  a'  tangent 
to  the  circle  in.  V,  and  the  #-trace  through  b  tangent  to  the 
circle  in  H. 


/  f 

V  1    i          a 

\       \      s'               / 

^      \  /           / 

\§S    I                      / 

1                              ' 

X% 

b            /  rx^ 

^, 

u                                                     \ 

p 

FIG.  47. 

Notes. — 1.  Four  planes  fulfilling  these  conditions  can  be  drawn,  all 
tangent  to  the  sphere. 

2.  Any  plane  parallel  to  GL  has  the  sum  of  its  angles  with  H  and  V  equal 
to  90°,  hence,  if  the  sum  of  the  given  angles  is  equal  to  90°,  the  problem  can 
be  easiest  solved  by  the  use  of  the  profile  plane. 

3.  If  it  be  required  to  pass  such  a  plane  through  a  given  point,  an  auxil- 
iary plane  may  be  drawn  anywhere,  as  in  the  foregoing,  filling  the  condi- 
tions, and  then  a  plane  may  be  passed  through  the  point  parallel  to  the 
auxiliary  plane.       The  plane  S,  Fig.  47,  is  passed  through  the  point  O, 
parallel  to  T  already  found.     The  plane  S  makes  the  same  angles  with  H 
and  V  that  T  does. 

Second  Method.  (From  Church  and  Bartlett.) — "If  a  line 
is  perpendicular  to  a  plane,  the  angles  between  the  line  and  the 
planes  of  projection  will  be  complements  of  the  angles  between 
the  plane  and  the  planes  of  projection.  Hence  if  we  construct 
a  line,  making  with  H  and  V  the  complements  of  the  required 
angles,  a  perpendicular  plane  through  this  point  will  make  the 
required  angles  with  H  and  V." 


80  PRACTICAL  DESCRIPTIVE  GEOMETRY 

99.  EXERCISES 

406.  Through   A(2  +  l$,  x)  B(3  +  i  y)  in  R(l+3)  1(4-2)  pass  a  plane  Q 
making  45°  with  R. 

407.  Draw  the  traces  of  a  plane  S,  making  60°  with  T(3  +  2)   5(1^-2), 
containing  the  line  C(3  +  2,  0)  D(l£,  0-2). 

408.  Draw  the  traces  of  a  plane  W,  making  30°  with  X(2  +  2)  4(2-1), 
through  D(3,  0-x)  E(3  +  y,  0)  in  X. 

409.  Draw  the  traces  of  a  plane  Q,   making  75°  with  R(5  +  2)   3(1-2), 
through  the  line  G(2,  0-l)H(5  +  2,  0). 

410.  Draw  the  traces  of  two  planes,  S  and  R,  each  making  30°  with  T(5  +  2) 
3(1-2),  containing  the  line  K(4  +  l,  0)  L(4,  0  +  1). 

411.  Draw  the  traces  of  two  planes,  W  and  Q,  each  making  45°  with  X 
(  +  2)oo(-li),  containing  the  line  M(2  +  2,  0)  N(3£,  0-1$). 

412.  Draw  the  traces  of  a  plane  Y,  making  60°  with  Q(  +  2)oo(-lJ),  con- 
taining the  line  O(3  +  2,  0)  P(3,  0-1|). 

413.  Draw  the  traces   of  a  plane  Q,  making  75°   with   R(2  +  2J)    (3|  +  2) 
(2-2J)  .(3i-2),  containing  the  line  A(2,  0-2J)  B(3£  +  2,  0).     What 
angle  does  Q  make  with  H  and  V? 

414.  Draw  the  traces   of  a  plane   S,   making  30°  with   T(2  +  2£)   (3^+2) 
(2-2i)  (3i-2),  containing  the  line  C(3i  +  2,  0)  D(3J,  0-2).     What 
angles  do  S  and  T  make  with  H? 

415.  Given  W(3  +  2)  5(1  £-2).     Pass  a  plane  X  that  will  make  45°  with  W 
and  cut  it  in  a  line  parallel  to  H,  and  1  in.  above  H. 

416.  The  plane  of  a  bin  cover  is  Y(  +  l)oo(  —  1^)  and  a  conveyor  belt,  16  in. 
wide,  touches  the  point  E(3  +  3  — 3),  and  runs  through  a  horizontal 
slot  in  the  bin  cover,  4  in.  X  18  in.,  at  an  angle  of  120°  with  Y.     Draw 
the  projections  of  the  slot.     Scale  f  in.  =1  ft. 

417.  Through  A(2  +  l-l)    B(4  +  3-2)   pass  a  plane  that  will  make  45° 
with  H.     What  angle  does  it  make  with  V? 

418.  Through  C(3  +  2  — i)  D(3  +  i-l£)  pass  two  planes  making  60°  with 
H,  and  measure  the  angle  between  them. 

419.  Through  E(2  +  1J  — f)  F(4  +  li-f)  pass  two  planes,  making  30°  with 
V.     What  is  the  angle  between  them? 

420.  Through  G(2  +  J-2)  H(4  +  2-$)  pass  two  planes,  making  75°  with 
//.     What  angles  do  they  make  with  V,  and  with  one  another? 

421.  A  hip  rafter  runs  from  M(2,  0-2)  to  N(4  +  2  — 1).    'Two  roofs,  each 
making  75°  with  H,  are  to  be  erected,  meeting  on  this  rafter.     If  the 
design  of  each  roof  is  an  isosceles  triangle,  what  will  be  their  H-  and  V- 
projections?     What  is  the  angle  between  the  roofs? 

422.  Draw  the  traces  of  a  plane  making  60°  with  V  and  30°  with  H,  con- 
taining the  point  M(3  +  1  —  1). 

423.  Draw  the  traces  of  a  plane  making  75°  with  V  and  45°  with  H,  con- 
taining the  point  A (4  +  1  —2). 

424.  Through  M(3  +  l  — 1)  draw  the  traces  of  all  the  planes  possible  that 
make  60°  with  V  and  45°  with  H. 

425.  Through  N(3  +  2  — 1)  draw  the  traces  of  a  plane,  making  30°  with  V 
and  75°  with  H,  using  the  second  method. 

426.  A  line  shaft  makes  15°  with  H  and  45°  with  V,  and  has  a  12-in.  pulley, 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       81 

6-in.  face,  with  its  center  at  O(3J  +  1—  2).     Draw  the  projections  of 
the  pulley.     Scale  1J  in.  =  1  ft. 

427.  A  man  is  reading  a  newspaper,  18  in.  X  24  in.,  and  his  eye,  at  P(2  +  2  —  1), 
is  directed  toward  the  center  of  the  paper,  2-ft.  distant,  at  an  angle  of 
30°  with  H  and  45°  with  V.  Draw  the  projection  of  the  newspaper  to 
a  scale  of  1  in.  =1  ft.  i 

100.  Problem  42.  To  find  the  shortest  distance  between  two 
lines  not  in  the  same  plane. 

Analysis. — 1.  Pass  a  plane  through  one  of  the  lines  parallel  to 
the  other. 

2.  Find  the  distance  from  any  point  in  the  second  line  to  this 
plane. 

This  will  be  the  shortest  distance  between  the  lines. 


1 


FIG.  48. 

Proof. — The  line  being  everywhere  equally  distant  from  the 
plane,  there  wall  be  one  point  where  a  perpendicular  from  the 
line  to  the  plane  would  intersect  the  line  in  the  plane.  This 
perpendicular  would  be  the  shortest  distance  between  the  lines, 
and  would  be  equal  to  any  perpendicular  let  fall  from  any  point 
in  the  line  to  the  plane. 

Let  the  student  make  and  explain  the  construction. 

101.  Problem  43.  To  find  the  shortest  line  that  can  be  drawn 
between  two  lines  not  in  the  same  plane. 


82  PRACTICAL  DESCRIPTIVE  GEOMETRY 

This  line  is  sometimes  called  the  "  common  perpendicular"  to 
the  two  lines. 

Analysis.  —  1.  Through  one  of  the  lines  pass  a  plane  parallel 
to  the  other  line. 

2.  Project  the  second  line  on  this  plane.       (Problem  29.) 

3.  Where  the  projection  of  the  second  line  intersects  the  first 
line,.  erect  a  perpendicular  to  the  plane,  intersecting  the  second 
line. 

This  will  be  the  required  line. 

Construction.  —  Let  it  be  required  to  find  the  shortest  line 
possible  to  be  drawn  between  the  lines  AB  and  CD  in  Fig.  48. 
1  1.  Through  any  point,  O,  of  AB,  draw  EF  parallel  to  CD. 

Note.  —  To  avoid  too  many  lines  in  the  figure,  the  projection  e'f  is  drawn 
to  coincide  with  c'd'.  This  is  not  absolutely  necessary,  however,  but 
convenient. 

2.  Pass  the  plane  T  through  AB  and  EF. 

3.  Project  CD  on  T,  in  the  line  MP. 

Note.  —  One  projection  only  need  be  found. 

4.  At  M,  where  MP  intersects  AB,  erect  a  perpendicular,  MN, 
to  the  plane  T,  intersecting  CD  at  N. 

The  required  line  is  MN.  Let  the"  student  prove  that  MN 
will  intersect  CD. 

102.  EXERCISES 

428.  Find  the  shortest  distance  from  A(lJ  +  2-2)   B(3  +  i~l)  to  0(5  +  1 
-1)  D(4  +  2-li). 

429.  Find  the  shortest  distance  from  E(l  +  3  -  2)  F(2  +  1  -  2)  to  G(3  +  1$  -  3) 


430.  Find  the  shortest  distance  from  K(2  +  1  +  1)  L(4  +  2  +  2)  to  M(3  -  \  -  £) 
N(4-2-2). 

431.  Find  the  shortest  distance  from  A  (2  +  1  -  2)  B(2  +  3-2)toC(3  +  2-l) 
D(5  +  l-3). 

432.  Find  the  shortest  line  parallel  to  H  that  can  be  drawn  between  E 
(l  +  i-3)  F(4  +  3-l)  and  G(2  +  3-2)  H(3i  +  l-2). 

433.  Find  the  common  perpendicular  to  K(l  +  l  —  1J)  L(3  +  3-3)  and  M 
(4  +  3-1)  N(5  +  l-3). 

434.  Find  the  common    perpendicular    to    A(l+2-3)     B(4  +  3-2)    and 


435.  Given  a  3-in.  cube  on  H.     Find  the  shortest  distance  between  the  non- 
intersecting  diagonals  of  the  base  and  one  of  the  adjacent  faces. 

436.  Find  the  shortest  line  that  can  be  drawn  from  E(2  +  1-3)F(4+1-1) 
to  G(2  +  3-l)  H(4  +  2-l). 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       83 

437.  Two  wires  run  from  K(l+3-3)  to  L(3  +  l-$)  and  from  M (2  +  3-1) 
to  N (5  + 1  —  2).     Find  where  the  shortest  connecting  wire  can  be  placed, 
and  its  length.     Scale  2  in.  =1  ft. 

438.  A  2-in.  pipe  is  run  from  a  hole  in  a  wall  at  O(3  +  3,  0)  45°  to  H  and  45° 
to  V.     A  1  J-in.  pipe  runs  from  P(l  +  3,  0)  in  the  wall  to  M,  2  ft.  straight 
out  from  the  wall.     It  then  is  turned  at  right  angles  and  runs  down- 
ward to  the  right  at  45°  to  H.     Where  can  the  shortest  connection  be 
made  between  the  two  pipes,  and  how  long  must  this  pipe  be  cut, 
allowing  2  in.  at  each  end  for  connections?     Scale  i  in.  =1  ft. 

439.  In  a  mill  there  is  a  bin,  a  6-ft.  cube,  placed  on  the  floor  against  the 
wall,  with  one  edge  at  A(12  ft.,  0,  0)  B(18  ft.,  0,  0).     A  24-in.  square 
chute  has  its  center  line  C(3ft.  +  9  ft.,  0)  D(16  ft.,  0-9  ft.)  running 
from  wall  to  floor.     It  is  desired  to  run  a  9-in.  cylindrical  pipe  from  a 
hole  at  E(18  ft.  +9  ft.,  0)  to  a  hole  in  the  floor  at  F(5  ft.  6  in.  +0  - 
14  ft.).     Can  the  pipe  pass  both  the  bin  and  the  chute?     How  much 
space  will  there  be  to  spare  in  each  case?     Scale  %  =  1  ft. 

440.  In   a  mill   room  there  are  two  pipes  with  center  lines  respectively 
A(l  +  3-l)    B(4,  0-1)  and  C(l,  0-1J)  D(4  +  2-l$).     Run  a  pipe 
from  E(3  +  2,  0)  in  the  wall  to  a  point  in  the  floor,  so  that  its  center 
line  will  be  exactly  6  in.  from  AB  and  CD.     Scale  1  in.  =1  ft. 

103.  The  Use  of  an  Auxiliary  Plane  of  Projection. 

Many  cases  arise  in  practical  work  and  in  abstract  problems 
where  an  auxiliary  plane  of  projection  may  be  used,  often  to  con- 
siderable advantage.  In  the  majority  of  cases  this  plane  is  taken 
perpendicular  to  H  and  oblique  to  V ,  yet  many  times  it  will  be 
drawn  perpendicular  to  V  and  oblique  to  H.  It  depends  on 
the  circumstances  which  decide  the  question.  Both  kinds 
will  be  used  in  our  illustrations. 

104.  Illustration  1. — To  find  the  line  of  intersection  between  a 
Hexagonal  Prism  in  natural  position  and  a  Triangular  Prism 
in  oblique  position. 

Fig.  49  shows  the  layout,  in  Third  Angle  projection,  of  such 
a  problem  with  the  lines  of  intersection  undetermined.  The 
surfaces  of  the  solids  are  entirely  made  up  of  planes,  and,  there- 
fore, the  lines  of  intersection  will  be  made  up  of  straight  lines. 
By  following  the  usual  method  of  procedure,  we  could  obtain 
the  various  lines  of  intersection,  (1)  by  obtaining  the  intersections 
of  the  various  planes,  each  with  each,  or  (2)  by  obtaining  the 
piercing  points  of  the  edges  of  one  prism  with  the  various  faces 
of  the  other,  and  then  connecting  these  piercing  points. 

In  many  cases  the  "usual"  method  is  difficult  or  cumber- 
some, and  the  use  of  the  auxiliary  plane  is  a  simple  and  quick 
method. 


84 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Construction. — The  auxiliary  plane  to  use  in  this  case  is  one 
parallel  to  the  bases  of  the  oblique  prism.  In  this  case  its  bases 
are  oblique  to  H  and  perpendicular  to  V.  Fig.  50  shows  the 


FIG.  49. 


FIG.  50. 


plane  Q  revolved  into  V  about  its  F-trace.     The  #-trace  revolves 

into  a  position  perpendicular  to  the  F-trace,  to  the  position  Qqt. 

The  two  solids  are  projected  on  Q  by  projecting  the  points 

perpendicularly  to  the  respective  traces  of  Q,  in  the  same  way 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES      85 

that  a  P-projection  would  be  made,  excepting  that  the  points 
are  rotated  through  an  acute  angle,  instead  of  90°. 

The  triangular  base  of  the  oblique  prism  appears  in  this  pro- 
jection in  its  true  size,  and  this  is  the  correct  way  of  obtaining 
its  H-  and  F-projections;  that  is,  by  drawing  it  first  on  Q,  and 
projecting  from  that  plane  to  H  and  V. 

The  intersecting  points  are  easily  located  from  the  Q-projection 
and  projected  back  to  the  surfaces  in  their  natural  position. 


I 


FIG.  51. 


105.  Illustration  2. — To  find  the  intersection  of  a  plane  and  a 
pyramid. 

Let  the  Hexagonal  Pyramid  and  the  Plane  T,  Fig.  51,  be  the 
given  surfaces. 

Discussion. — If  a  plane  Q  perpendicular  to  the  //-trace  of  T 
be  employed  as  an  auxiliary  plane  of  projection,  it  will  also  be 
perpendicular  to  H,  and  may  be  regarded  as  a  vertical  plane  of 
projection  with  its  H-trace  qq  as  the  ground  line.  The  new 


80  PRACTICAL  DESCRIPTIVE  GEOMETK 

ground  line  may  be  placed  at  any  distance,  or  on  either  side 
of  the  //-projection,  wherever  there  is  a  good  open  space.  The 
advantage  of  this  arrangement  is  that  T  is  a  projecting  plane 
on  Q7  and  consequently  all  points  in  T  will  be  projected  on  Q 
in  its  Q-trace. 

Analysis. — 1.  Draw  the  #-trace  of  Q,  perpendicular  to  the 
#-trace  of  T. 

2.  Project  the  pyramid  on  Q. 

Note. — All  points  projected  on  Q  are  the  same  distance  from  qq  that 
the  F-projections  are  from  GL. 

3.  Obtain  the  Q-trace  of  T. 

Note. — If  the  intersection  of  the  Q  and  T  vertical  traces  is  outside  the 
limits  of  the  problem,  an  auxiliary  plane  X  may  be  passed  parallel  to  Q, 
and  the  Q-trace  of  T  will  be  parallel  to  the  X-trace  of  T,  see  XxY  The 
Q-trace  of  T  can  then  be  drawn. 

4.  The  points  a1?  b1?  c1;  etc.,  will  be  the  piercing  points  of  the 
edges  of  the  pyramid  with  T. 

5.  Project   these   points,    first   to    the   //-projections    of   the 
respective    edges,    and    thence    to    their    F-projections.     The 
intersection  will  then  be  the  cross-hatched  figures. 

Note. — In  case  of  difficulty  in  locating  exact  intersections  on  the  /^-pro- 
jection, owing  to  the  parallelism  or  "near-parallelism"  of  the  projectors 
with  the  lines  on  H,  it  will  be  easy  to  locate  them  first  on  the  F-projection, 
because  all  points  projected  on  Q  are  the  same  distance  from  qq  that  their 
F-projections  are  from  GL. 

The  student  may  argue  that  the  foregoing  illustrations  are  more  trouble 
than  they  are  worth.  This  may  be  true  in  the  case  of  simple  problems,  such 
as  these  two  are,  but  many  difficult  cases  will  arise  where  this  method  will 
be  found  a  great  saving.  In  all  cases  this  method  is  more  exact  than  the 
plodding  method. 

106.  PRACTICAL  EXERCISES 

441.  Find  the  length  of  the  various  rafters,  MN,  NK,  etc.,  in  the  roof,  shown 
in  Fig.  52,  and  the  angles  they  make  with  the  horizontal.     Scale 
lin.=l  ft. 

442.  Measure  the  angles  between  the  rafters  MN,  NK,  etc.,  and  the  ridges 
that  intersect  them.     (Fig.  52.) 

443.  Measure  all  the  valley  angles  between  the  planes  A  and  C,  E  and  G,  etc., 
in  the  roof.     (Fig.  52.) 

444.  A  rectangular  chimney  3  ft.  6  in.  X5  ft.  4  in.  is  to  be  put  through  the 
roof  C  in  Fig.  52.     Find  the  H-  and  F-projections  and  true  size  of  the 
hole  in  the  roof. 

445.  In  the  center  of  roof  B  in  Fig.  52,  a  rectangular  skylight  3  ft.  6  in.  X 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       87 


4  ft.  6  in.  is  to  be  built.  Find  its  H-  and  F-projections.  In  the  center 
of  roof  H,  a  4-ft.  square  opening  is  to  be  cut.  Draw  its  projections. 
446.  An  octagonal  stone  tower,  20  ft.  in  diameter  and  20  ft.  high,  is  to  be 
built,  surmounted  by  a  pyramidal  cap,  12  ft.  high.  Draw  its  pro- 
jections, and  draw  in  the  center  of  each  of  the  four  visible  faces  of  the 
cap  one  of  the  styles  of  windows  shown  in  Fig.  53.  Draw  only  one-half 


FIG.  52. 


Scale 


447. 


of  the  plan,  and  part  of  the  elevation  of  the  tower  and  cap. 
i  in.  =  1  ft. 

A  12-in.  steam  feed  pipe  runs  through  a  shed,  its  center  line  being 
A(l  +  li-3)  B(5  +  li-i).  The  plane  of  the  shed  roof  is  T(5  +  3) 
1(5  —  2).  Find  the  projections  and  true  size  of  the  opening  for  the  pipe. 
Scale  1  in.  =  1  ft. 

448.  A  trapeze  bar,  3  ft.  long,  is  suspended  by  ropes  3  ft.  long.     Find  how 
high  it  is  raised  by  twisting  the  trapeze  through  90°. 

b.  c. 


Windows  2'0"x3'6': 
FIG.  53. 

449.  A  conical  tower  is  20  ft.  in  diameter  and  20  ft.  high.     Draw  two  lines 
from  the  apex  to  the  base,  making  45°  with  one  another.     Scale  |  in.  = 
1ft. 

450.  A  cube  of  \\"  sides  is  erected  on  H  with  its  base  diagonal  ||  to  V. 
Draw  the   traces    of   a  plane  Q  1 1  to  GL,  which  shall  cut  a  regular 
hexagon  from  the   cube.     Revolve  the   hexagon  to  prove  that  it  is 
regular  and  calculate  the  angle  that  Q  makes  with  H. 


88 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


451.  A  square  building  is  covered  by  a  hip  roof,  which  forms  a  regular 
pyramid.     The  hip  rafters  make  30°  with  H.     What  angle  do  the 
roof  planes  make  with  H?     What  angle  do  two  adjacent  roof  planes 
make  with  one  another?     Use  any  convenient  scale. 

452.  A  3-in.  cube  is  drilled  at  two  points,  A  on  the  top  face  f  in.  from  both 
the  right  and  rear  edges,  and  B  on  the  left  side  face  f  in.  from  the 
lower  and  front  edges.     The  holes  are  drilled  straight  in  at  A  and  B 
respectively  1£  in.  and  l\  in.  deep.     At  what  point  on  the  front  face, 
and   at  what  angles  with  H  and  V  must  a  hole  be  drilled  to  join  the 
bottoms  of  the  holes?     Full  size. 

453.  Two   pulleys,   24  in.  X8  in.,   are   mounted   on   parallel   shafts.     The 
center  line  of  one  shaft  is  A(l  +  2-3)   B(2£  +  2-l),  and  the  other 
center  line  passes  through  C(5  +  l  — 1^).     The  center  of  the  pulley 
on  AB  is  at  B.     The  belt  passes  through  a  wall  whose  plane  is  T(5  +  3) 
3(1  —  1).     Make  a  rectangular  opening  in  the  wall  to  accommodate 
both  belts   (disregarding  clearance).     Draw  its  projections  and  give 
its  true  size.     Draw  to  a  scale  of  \  in.  =  1  ft. 

454.  Inscribe  a  sphere  in  the  tetrahedron  A(2,  0-1)  B(4,  0-1)  C(3J,  0-3) 


A. 


D. 


Moulding      Sections. 
FIG.  54. 


455.  Inscribe    a     sphere     in    the    tetrahedron    E(l£  +  l-2),    F(2  +  1-J) 
G(4  +  2-2)  H(3  +  i-l). 

456.  Circumscribe  a  sphere  about  the  tetrahedron  M(l,  0-2)  N(3,  0-1) 
0(4,  0-2i),  P(2  +  li-li). 

457.  Find  the  projections  of  the  circle  passing  through  the  points  A(2  +  1  —  1), 
B(3  +  li-i),  and  C(4  +  i~2). 

458.  Pass  a  plane  equidistant  from  D(l+3  —  1)  E(3  +  J-3)   and  F(4  +  J 


459.  Draw  the  H-  and  F-projections  of  a  If  in.  hexagonal  nut,  chamfered 
and   drilled   but   not    threaded,    having    its   center   lir>e   K(2  +  2  —  1) 
L(4  +  l-2).     Scale  full  size. 

460.  M(l  +  i-i)  N(4  +  3-3)  and   O(3  +  3-2)  P(5|,  0-1)  are  two  pipes. 
What  is  the  length  of  the  shortest  connection  that  can  be  made  be- 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       89 

tween  them,  parallel  to  H,  allowing  2  in.  for  the  connection  ends. 
Scale  1  in.  =  1  ft. 

461.  Determine  the  outline  of  a  moulding  cutter,  mounted  on  an  arbor  of 
3  in.  diameter,  to  cut  moulding  of  one  of  the  cross-sections  shown  in 
Fig.  54.     Let  the  center  line  of  the  arbor  be  3  in.  from  the  nearest 
point  of  the  moulding.     Full  size. 

462.  Erect  a  pyramid  on  a  2-in.  square  in  H,  making  the  angle  between 
two  adjacent  oblique  faces  120°. 

463.  At  a  point  of  outcrop  A(4  +  2  —  1)  the  strike  of  a  body  of  ore  is  South, 
50°  West.     The  dip  is  45°.     A  tunnel  is  driven  from   B(l+3-f), 
which  is  on  the  surface,  to  the  ore  perpendicularly.     Where  will  it 
strike  the  ore? 

Note. — The  "strike"  of  a  stratum  of  ore  is  a  horizontal  line  in  the  plane 
of  the  ore. 

The  "dip"  is  the  inclination  of  the  ore  plane  to  the  horizontal. 

464.  The  point  of  outcrop  is  A(200  ft.  +  200  ft.  -  100  ft.),  and  a  line  is  run  to 
B,  a  distance  from  A  200  ft.,  South,  30°  West.     The  altitude  at  B  is 
125  ft.,  and  a  vertical  bore  hole  strikes  ore  at  a  depth  of  25  ft.    Another 
bore  hole  is  made  at  C,  on  the  same  level  as  B,  and  150  ft.  distant 
from  A  and  B.     At  C  the  hole  is  drilled  100  ft.,  striking  the  ore. 
Determine  the  dip  and  strike  of  the  vein.     Scale  1  in.  =100  ft. 

465.  Points  D,  E  and  F  are  on  the  side  hill,  which  dips  20°  East.     D  is  a 
point  of  outcrop  of  a  vein  of  ore.     From  D,  E  bears  South  45°  East 
400  ft.  on  the  slope  of  the  hill,  and  F  bears  South  15°  East  600  ft. 
on  the  slope  of  the  hill.     Vertical  drill  holes  at  E  and  F  strike  ore  at 
100  ft.  and  150  ft.  respectively.     Determine  the  dip  and  strike  of  the 
vein.     Scale  1  in.  =  100  ft. 

466.  At  A(8  ft, +24  ft.,  0)  and  B(20  ft.  +  24ft.,  0)  in   a   mill  wall,  two 
parallel  6-in.  pipes  run  downward  to  the  right  at  45°  to  the  floor  and 
30°  to  the  wall.     It  is  proposed  to  run  a  12-in.  cylindrical  chute  on 
the  center  line  C(30  ft. +  16  ft.,  0)  D(4  ft.,  0-24  ft.).     Can  this  be 
done?     If  not,  how  much  could  D  be  shifted  toward  the  wall  to  effect  it? 
Scale  |  in.  =  1  ft. 

468.  A  pulley,  24  in.  diameter,  and  12  in.  face,  runs  on  the  shaft  A(2  ft. 
+  2  ft. -6  ft.)  B(10ft.  +  7  ft. -2  ft.).     Draw  the  projections  of  the 
pulley.     Scale  %  in.  =  1  ft. 

469.  Draw  the  projections  of  a  l|-in.  sphere  tangent  to  the  plane  T(l+3) 
4(1-2)  at  the  point  O(3  +  i,  x)  in  T. 

470.  Measure  the  various  lineal  and  dihedral  angles  of  a  hexagonal  pyr- 
amid of  2  in.  base  and  1\  in.  altitude. 

471.  One  line  of  a  roof  is  A(2,  0-2)  B(4,  0- J).     At  A  a  rafter  runs  per- 
pendicularly to   AB  and  60°  to  H  in  a  northwesterly  direction.     At 
B  the  rafter  is  parallel  to  V,  and  the  ridge  is  horizontal,  2  in.  (scale) 
above  H.      Draw  both  projections  of   the  roof,  and  find   the  traces 
of  its  plan".     Compute  its  area. 

472.  Draw  the  projections  of  a  10-ft.  circular  skylight  in  a  roof  whose  plane 
isS(8ft.  +  24  ft.)  32  ft.  (8  ft. +  16  ft.),  with  its  center  at  M(20ft.+ 
8  ft.,  x)  in  the  roof.     Scale  J  in.  =  1  ft. 


90 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


473.  A  tunnel  through  a  hill  runs  from  A(425  ft. +275  ft.  — 100  ft.)   to 
B(2100  ft.  + 150  ft.  - 1300  ft.).     Find  its  length  and  grade. 

474.  A  plane  T  passes  through  GL  and  the  point  K(3  +  l  — 2).     A  plane  S 
parallel  to  it  cuts  V  1£  in.  above  GL.     Find  the  //-trace  of  S,  the 
distance  between  T  and  S,  and  their  angles  with  H  and  V. 

475.  Pass  a  plane  through  L(l+2-l)  equally  distant  from  M(4  +  l-f), 
N(3  +  2  — £),  and  O(2  +  2-|).     Find  a  point  in  the  plane  thus  found 
which  is  equally  distant  from  M,  N  and  O. 

476.  Erect  a  cube  on  H  with  its  base  diagonal  A(3,  0-$)  B(4,  0-1±). 
Show  its  picture  as  seen  from  C(3J  +  lf  — 3).     Show  the  picture  in 
its  true  size. 

Note. — Pass  the  picture  plane  perpendicular  to  the  line  from  the  point  of 
sight  to  the  center  of  the  cube. 


FIG.  55. — Mission  lamp. 


477.  From  a  hole,  A(18  ft. +  10  ft.,  0),  in  a  mill  wall,  run  a  pipe  that  will 
make  a  straight  line  connection  with  two  pipes,  whose  center  lines  are 
B(3  ft.,  0-8  ft.)  C(10  ft. +  10  ft. -18  ft.  6  in.)  and  D(3  ft. +  10  ft.- 
5  ft.)  E(10  ft.,  0-14  ft.).     What  are  the  lengths  of  the  two  portions  of 
pipe,  making  no  allowance  for  their  connections?     Scale  \  in.  =  1  ft. 

478.  Draw  the  projections  of  the  circle  passed  through  A(2  +  2-J),  B(2£  + 
1-1),  and  C(3i  +  2-li). 

479.  The  hinge  of  a  door  is  M(4  +  3,  0)N(4,  0,  0).     When  shut,  it  is  to  the 
left  of  MN.     A  taut  wire  runs  from  O(3  +  2,  0)  to  P(5,  0-1).     Find 
the  outline  of  a  slot  that  will  allow  the  door  to  open  freely  through  90° 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       91 

480.  Measure  the  angles  that  the  planes  A  and  B,  of  the  lampshade,  shown 
in  Fig.  55,  make  with  one  another,  and  with  H  and  V. 

481.  Make  a  full  set  of  details  to  scale  of  all  the  glass  and  sheet  iron  pieces  in 
the  lamp  shade  and  supports,  shown  in  Fig.  55.     Do  not  duplicate 
the  details,  but  state  how  many  are  required  of  each.     Allow  for  the 
folding  of  the  iron  pieces. 

482.  Make  a  detail  of  the  backbone  of  the  saw  horse  shown  in  Fig.  56, 
showing  the  angles  of  the  cuts  to  be  made.     Draw  to  any  convenient 
scale. 

483.  Make  detail  drawings  of  the  legs  and  braces  of  the  saw  horse,  shown  in 
Fig.  56,  showing  the  angles  of  the  various  cuts.     Draw  to  any  con- 
venient scale. 


FIG.  56.— Saw  horse. 


484.  Make  a  pattern  to  any  convenient  scale  of  the  grain  hopper,  shown  in 
Fig.  57.     Use  two  spaces  for  the  projections  and  pattern.     Make  a 
paper  model  of  the  hopper,  putting  on  a  flap  for  holding  together. 

485.  Make  a  pattern  to  any  convenient  scale,  of  the  ventilator  cap,  shown 
in  Fig.  58. 

486.  Let  H  be  a  mirror.     A  ray  of  light  from  A(l-+2  — 2)  passes  through 
B  (3  +  \  - 1) .     At  what  point  does  it  strike  Vf 

487.  Find  the  path  of  a  ray  of  light  that  passes  through  M(2  +  3-2)  and  is 
reflected  from  H  and  V  to  N(4i  +  2-$). 

488.  The  plane  of  the  top  of  a  bin  door  is  T(l+2)3$(l-3).     The  center 
line  of  a  12-in.  square  chute  to  the  bin  runs  from  O(4  +  3,  0)  runs  at  an 
angle  of  30°  to  H  and  30°  to  V.     Find  the  opening  in  the  bin,  and  its 
true  size.     Scale  to  suit. 


92 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


489.  Fig.  59  shows  a  stool,  or  plant  stand,  made  of  1-in.    lumber.     The 
edges  of  the  four  inclined  supporting  pieces  are  chamfered   to  fit. 
Make  a  working  drawing  with  full  details  to  scale,  showing  the  angles 
of  the  champered  pieces,  without  duplicating  any  of  the  parts.     Take 
two  spaces  for  the  work. 

490.  Draw  the  projections  of  four  spheres  of  1  in.,  1£  in.,  1|  in.,  and  If  in. 
respective  diameters,  each  sphere  tangent  to  the  other  three. 


ip 


FIG.  57. — Grain  hopper. 


FIG.  58. — Ventilator  cap 


\ 

/ 

7 

1 

TN                   '"I 

1 
1 

1 

ii                   M 
M                   i  i 
i  j-                 j  i 

1 

I 

/ 

/ 

\ 

- 


X- 


18' -H 

14-" »    I 


FIG.  59.— Stand. 


491.  Develop  the  surface  of  the  keystone  of  the  arch  shown  in  Fig.  60.     The 
masonry  is  12  in.  thick.     Use  any  convenient  scale. 

492.  A   borehole   cuts   a  6-ft.    core   from  a    vein,    whose    strike   is   North 
60°  East,  and  whose  dip  is  45°  to  the  Northwest.     The  borehole  bears 
South  75°  West  and  dips  30°.     What  is  the  real  thickness  of  the  vein? 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       93 

493.  At  A,  where  the  elevation  is  9800  ft.,  a  vertical  borehole  strikes  a  vein 
at  300  ft.     B  bears  North,  30°  East,   1500  ft.  from  A,  and  has  an 
elevation  of  9600  ft.     An  inclined  hole  is  drilled,  bearing  North  75° 
East,  dipping  75°,  striking  the  vein  at  525  ft.     At  C,  whose  elevation 
is  9900  ft.,  1200  ft.  from  A,  South  60°  East,  a  vertical  bore  hole  cuts 
the  ore  at  375  ft.     Required  the  dip  and  strike  of  the  vein.     Let  H 
be  at  9500  ft.  level. 

494.  At  A(l+3-3)   a  borehole  is  drilled,   bearing  North  30°  East,    and 
dipping  75°.     It  intersects  a  vein,   whose  strike  is  due  North,  and 
cuts  an  8-ft.  core.     The  true  thickness  of  the  vein  is  5  ft.     What  is 
the  dip  of  the  vein? 

495.  Vertical  holes  on  a  hill  side  (considered  a  plane)  are  bored  150  ft.  at 
A(100  ft.  +  400  ft.-300  ft.),  200ft.  at  B(350  ft. +  250  ft.-325  ft.), 
and  125  ft.  at  C(600  ft. +  325  ft. -100  ft.),  striking  a  vein  of    ore. 
What  is  the  line  of  outcrop,  and  what  are  the  dip  and  strike  of  the 
vein? 


FIG.  60. — Stone  arch. 


496.  Find  the  line  of  intersection  between  the  square  pyramid  A(3,  0  — f) 
B(4i,  0-2)   C(3,  0-3i)   D(lf,  0-2)    E(3+ 2f-2),  and  a  triangular 
prism,  its  lower  face  parallel  to  H,  its  center  line  M(2  + 1  —  1)    N(5+l 
—  2),  and  its  cross-section  an  equilateral  triangle  of  1-in.  sides. 

497.  Find  the  line  of  intersection  between  a  hexagonal  prism  of  2-in.  base 
and  2^-in.  altitude,  standing  on  H,  and  a  1-in.  square  prism,  whose 
axis  is  parallel  to  H,  1  in.  above  H  and  j  in.  away  from  the  center  line 
of  the  hexagonal  prism. 

498.  Erect  a  square  prism  with  its  base  on  H,  of  1^-in.  base  and  2-in.  alti- 
tude.    Make  the  base  diagonal  parallel  to  V.     Intersect  by  a  hexag- 
onal prism,  lj  in.  long  diameter,  with  two  faces  parallel  to  V.     Let 
the  two  center  lines  intersect  at  an  angle  of  60°  at  a  point  1  in.  above 
H.     Find  the  line  of  intersection. 

499.  Erect  a   2-in.    cube    on    H.     Draw  a   1-in.   triangular  prism,   whose 
center  line  passes  through  the  upper,  rear,  right-hand  corner  of  the 
cube,  making  angles  of  45°  with  H  and  30°  with  V,  running  through 
the  cube.     Find  the  intersection  of  the  solids. 

500.  A(2  +  l-2)     B(4  +  2-l)  is  the  center  line  of  a  4  in.Xl2  in.  hollow 
tile,  of  square  section  with  rounded  corners.     The  walls  are  \  in.  thick, 


94  PRACTICAL  DESCRIPTIVE  GEOMETRY 

and  the  radius  of  the  outside  rounded  corners  is  1  in.     Draw  its  pro- 
jections quarter  size. 

SHADES  AND  SHADOWS 

107.  One  of  the  most  interesting  applications  of  the  elementary 
work  in  Descriptive  Geometry  is  found  in  obtaining  the  shades 
and  shadows  of  objects.  Naturally,  this  application  makes  its 
greatest  appeal  to  the  architectural  student,  but  no  engineer 
should  be  without  knowledge  of  its  workings,  although  he  may 
rarely  make  professional  use  of  it. 

The  distinction  between  Shades  and  Shadows  is  determined  by 
their  definitions. 

Shade  is  that  part  of  an  object  not  exposed  to  the  direct  rays 
of  light. 

Shadow  is  that  portion  of  a  surface  in  light  from  which  rays 
of  light  are  excluded. 

Umbra  is  the  unlighted  space  behind  an  object  in  light. 

Note. — Shadow  is  the  intersection  of  the  umbra  with  a  surface  in  light. 
That  is,  it  is  the  interrupted  portion  of  an  umbra  by  an  interposed  surface. 


FIG.  61. 

The  Line  of  Shade  is  the  imaginary  line  between  the  light  and 
shaded  portions  of  an  object. 

The  picture  (Fig.  61)  shows  the  distinctions  between  these 
characteristics  of  light  effects. 

The  sun  is  our  source  of  light,  and  its  rays  are  assumed  to  be 
straight  lines  all  parallel.  This  assumption  is  very  nearly  true, 
and  the  deviation  of  the  rays  from  parallelism  is  so  very  slight 
that  no  account  need  be  taken  of  the  error.  For  convenience  in 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       95 

conventional  shading,  the  sun's  rays  are  always  assumed  to  be 
passing  over  the  left  shoulder  of  the  observer  in  the  direction  of 
the  diagonal  of  a  conventional  cube;  that  is,  both  the  H  and  V 
projections  of  the  rays  make  45°  with  GL. 

Referring  to  Fig.  61,  we  have  the  rays  of  light  depicted  as 
coming  toward  a  sphere  in  parallel  lines,  and  either  stopping,  if 
they  strike  the  sphere,  or  passing  on,  if  they  do  not.  One-half 
of  the  sphere  is  in  light,  and  one-half  in  shade,  and  the  line  of 
shade  is  a  great  circle  of  the  sphere.  This  determines  the  umbra 
to  be  a  cylinder  of  diameter  equal  to  the  diameter  of  the  sphere. 
The  umbra,  on  meeting  another  surface,  is  interrupted,  and  its 
intersection  with  that  surface  is  the  shadow. 

108.     Shadows  made  up  of  piercing  points  of  rays. 

From  the  foregoing,  it  will  appear  that  the  integral  of  the 
tangent  rays  (the  tangent  surface)  to  the  sphere  will  pierce  the 
surface  in  light  in  the  outline  of  the  shadow.     Hence,  the  finding 
of  the  line  of  shade  resolves  itself  into 
finding  the  tangent  rays  of  the  object, 
their  points  of  tangency  making  up  the 
line  of  shade.     These  tangent  lines  form 
the   surface    of   the   umbra,    and   their 
piercing  points  with  the  surface  in  light  \p, 

form  the  outline  of  the  shadow. 


The  following  rules  are  thus  deduced: 

109.  The  Shadow  of  a  Point  is  a  Point. 
The  Umbra  of  a  Point  is  a  Line. 

110.  The  Shadow  of  a  Line  is,  in  gen- 
eral, a  Line.  ° 

The  Umbra  of  a  Line  is,  in  gen-         The  shadow  oTa  point, 
eral,  a  Plane. 

111.  The  Shadow  of  a  Plane  Figure,  cast  on  a  plane,  is,  in  gen- 
eral, a  Plane  Figure  of  similar  outline. 

The  Umbra  of  a  Plane  Figure  is,  in  general,  a  Prism. 

112.  The  Method  of  Finding  Shadows. 

Fig.  62  shows  a  point  O  in  space,  acted  upon  by  a  ray  of  light 
in  the  conventional  direction,  OP.  The  F-piercing  point  of  the  ray 
OP  is  P,  which  is  the  shadow  of  O  on  V.  Fig.  63  shows  a  line 
AB  in  space.  From  A  and  B  draw  rays  of  light  AC  and  AD, 
having  their  H-  and  F-projections  45°  from  GL.  Find  their 
//-piercing  points,  C  and  D.  The  shadow,  therefore,  of  AB  on 


96 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


H  is  CD.     The  umbra  of  the  line  is  the  plane  ABCD,  and  its 
#-trace  is  the  shadow,  CD. 

Note.— In  case  one  end  of  the  line  casts  its  shadow  on  H,  and  the  other 
end  casts  its  shadow  on  V,  the  two  traces  of  the  umbra  plane  will  be  the 
shadow,  and  it  will  be  a  broken  line,  partly  on  H,  and  partly  on  F.  This 
condition  will  be  found  later  in  Fig.  66. 


FIG.  63. — The  shadow  of  a  line. 

a'  b'  c1  of 


.e'f 


-719 


fL- 


FIG.  64. — The  shadow  of  a  plane  figure. 


To  find  the  shadow  of  a  square  (or  other  plane  figure) . 
Let  ABCD  (Fig.  64)  be  a  square,  and  let   AE,  BF,  CG   and 
DH  be  the  rays  passing  the  four  corners.     The  H  -piercing  points 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       97 

of  the  four  rays  are  E,  F,  G  and  H,  and  the  quadrilateral  formed 
by  joining  them  in  order  is  the  shadow  of  the  square  on  H. 

Note. — If  the  square  is  parallel  to  H  (as  in  this  figure)  the  shadow  will  be 
an  equal  square.     Prove  this  statement. 

To  obtain  the  shadow  of  a  cube  (or  other  solid  figure). 
Draw  the  projections  of  a  cube,  such  as  is  shown  in  Fig.  65. 
From  B,  C  and  D  draw  rays  of   light,  piercing  H  in  the  points 
E,  F  and  G. 


FIG.  65.— The  shadow  of  a  cube. 

The  line  of  shade  follows  this  course:  MBCDN.  The  line  MB 
casts  the  shadow  ME,  the  line  BC  casts  the  shadow  EF,  the  line 
CD  casts  the  shadow  FG,  and  the  line  DN  casts  the  shadow  NG. 
This  gives  us  the  entire  space,  mefgncm.  as  the  shadow  of  the  cube 
on  H.  No  ray  runs  past  A,  because  it  is  on  the  light  surface. 
A  ray  through  A  would  penetrate  the  cube,  and  would  merely 
be  inside  the  umbra,  and  would  pierce  the  surface  in  light  within 


98 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


the  shadow,  and  not  on  the  outline.  In  shadows,  only  points  on 
the  outline  are  necessary  to  be  found. 

To  obtain  the  shadow  of  an  object  when  it  falls  on  more  than 
one  plane. 

Let  it  be  required  to  draw  the  shadow  of  the  hexagonal  prism 
shown  in  Fig.  66. 

The  procedure  in  this  case,  as  in  all  cases,  is  to  draw  rays 
from  the  corners  of  the  object,  and  find  their  piercing  points 
with  the  first  object  they  strike. 


FIG.  66. — The  shadow  of  a  solid  falling  on  two  planes. 

In  this  figure  all  the  rays  but  that  from  X  pierce  the  F-planc 
first,  yet  the  object  is  some  distance  from  V,  and,  therefore,  a 
portion  of  the  shadow  must  be  on  H. 

The  ray  from  A  pierces  V  at  C,  but  the  shadow  of  the  line  AB 
is  partly  on  V  and  partly  on  H.  The  shadow  of  AB  is  drawn 
on  H  until  it  intersects  GL,  then  from  that  point  it  is  joined  with 
the  F-piercing  point  of  AC.  Thence  the  outline  is  drawn  through 
the  F-piercing  points  of  the  rays  from  D  and  M,  respectively  E 
and  N.  The  line  MX  casts  a  shadow  on  both  V  and  H.  The 
piercing  point  of  the  M  ray  is  at  n'  and  that  of  the  X  ray  is 
at  y.  The  shadow  of  MX  will  be  a  broken  line.  Taking  any 
intermediate  point  on  the  line,  and  drawing  from  it  a  light  ray, 
and  finding  its  piercing  point,  will  determine  the  direction  of 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES       99 


its  shadow  on  either  plane.  The  easiest  way  is  to  find  the  H- 
piercing  point  of  the  M  ray  at  O.  The  shadow  of  MX  on  H  will 
then  be  the  line  yo.  From  the  point  where  yo  crosses  GL,  the 
shadow  is  interrupted  by  V,  so  connect  that  point  with  n',  and 
the  shadow  is  complete. 

The  foregoing  illustrations  furnish  the  rudiments  of  "  Shades 
and  Shadows."  By  working  the  following  exercises,  you  will 
obtain  a  fair  knowledge  of  the  subject,  as  the  most  complicated 
work  of  the  kind  is  merely  an  elaboration  of  these  principles. 

EXERCISES 

113.  In  these  exercises  use  double-cross-hatching  for  the  shadows,  and 
single  for  the  shades,  and  dotted  lines  for  the  invisible  part  of  the  shadow, 
as  in  Fig.  66.  ' 

501.  Erect  a  cube  on  H,  having  A(3,  0-1)    B(3,  0-3)  for  its  base  diagonal. 
Determine  its  shades  and  shadows. 


^e- 


-x- 


FIG.  67. 


502. 


Find   the   shades   and    shadows   of  an   oblique   cone,    whose  axis   is 
C(2  +  0-2)  D(4  +  2-l),  and  whose  base  is  a  l£in.  circle  in  H. 
Let  E(3,  0-1J)   F(3  +  3-li)  be  the  center  line  of  a  hexagonal  pyra- 
mid of  2  in.  base.     Find  its  shades  and  shadows. 

Let  G'(  1  i  + 1  - 1  i )  H  (4  +  f  -  2)  be  the  center  line  of  a  hexagonal  prism, 
with  one  face  on  //.     Draw  its  shades  and  shadows. 

505.  Find   the   shades   and    shadows    of   a   2-in.  cylinder,    whose    axis"  is 
M(2,0-2)  N(2  +  3-2). 


503. 


504. 


100 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


506.  Find  the  shades  and  shadows  of  a  cylinder  lying  on  H,  whose  axis  is 
0(li  +  f-l)  P(3  +  |-2). 

507.  Find  the  shadow  on  H  and  T(5  +  2)l(5-2)  of  a  1  in.  square  prism, 
2  in.  high,  having  its  base  center  at  A(2,  0  —  2). 

508.  Find  the  shadow  on  H  and  T(5  +  3)3(3-3)  of  a  hexagonal  pyramid, 
of  1£  in.  base  and  2  in.  altitude,  the  center  of  its  base  being  B(2,  0  —  2). 

509.  Find  the  shades  and  shadows  of  a  1  in.  square  prism,  2  in.  high,  sur- 
mounted by  a  1£  in.  square  plinth,  ^  in.  thick.     The  center  line  of 
both  solids  is  A(l$,  0-2)  B(lJ  +  2i-2). 

510.  Make  the  prism  in  Exercise  509  a  1  in.  cylinder,  and  the  exercise  other- 
wise the  same. 

511.  Find  the  shades  and  shadows  of  a  cross  with  the  dimensions  and  posi- 
tion shown  in  Fig.  67.     Place  the  center  of  the  base  at  A(2,  0  —  1). 


CXJ 


2"- 


FIG.  68. 


512.  Place  the  cross  having  the  same  dimensions  and  position  as  that  given 
in  Fig.  67,  with  its  center  at  B(2,  0-2).     Find  its  shades,  and  its 
shadow  on  H  and  T(4  +  2)  1  (4  -  2) . 

513.  Draw  a  stepped  pyramid,  made  up  of  four  plinths,  each  £  in.  thick, 
and  If  in.,  1^  in.,  1^  in.,  and  1  in.  square  respectively.     Draw  the 
cross    of    dimensions  and   position   shown  in  Fig.  67  on   the   top   of 
the  pyramid.     Obtain  the  shades  and  shadows.     Draw  the  stepped 
pyramid  well  to  the  left  of  the  drawing,  and  "square"  with  V '. 

514.  Place  a  prism,  whose  right  section  is  a  1^  in.  equilateral  triangle,  1|  in. 
long,  on»a  1  in.  cube  (so  as  to  resemble  a  house  and  roof),  and  find 
their  shades  and  shadows. 

515.  Draw  the  box  (Fig.  68)  so  that  one  of  its  long  sides  is  30°  from  V,  and 
draw  its  shades  and  shadows,  inside  and  outside,  and  on  H  and  V. 

516.  Place  the  box  (Fig.  68)   on  the  plane  8(1  +  1)3(3-3),  and  find  its 
shades  and  shadows,  inside  and  outside,  on  H,  V  and  S. 

517.  Place  the  box  (Fig.  68)  on  end  with  the  bottom  toward  and  parallel 
to  V,  and  draw  its  shades  and  shadows. 


CHAPTER  III 
CURVED  LINES  AND  SURFACES 

114.  Definitions. 

A  Line  is  the  path  of  a  point. 

A  Straight  Line  is  a  line  in  which  the  point  moves  continually 
in  .one  direction. 

A  Curved  Line  is  a  line  in  which  the  point  moves  in  a  direction 
that  is  being  changed  constantly  in  obedience  to  some  law. 

A  line  is  regarded  as  having  one  dimension  only — length, 
without  breadth  or  thickness.  It  is  considered  as  being  made 
up  of  infinitely  small  points,  separated  by  infinitely  small  spaces. 
The  line  may,  therefore,  be  regarded  as  being  generated  by  an 
immaterial  point,  occupying  consecutive  positions.  Two  of  these 
consecutive  points  determine  a  straight  line,  of  infinitesimal 
length,  but  of  a  determinate  direction.  Such  a  line  is  called  an 
Element;  that  is,  it  is  one  of  the  innumerable  infinitely  small 
straight  lines  that  compose  every  line,  straight  or  curved. 

115.  Classification  of  Lines. 
Lines  are  Straight  or  Curved. 

Curved  Lines  may  be  Plane  Curves,  also  called  Lines  of  Single 
Curvature,  or  they  may  be  Space  Curves,  also  called  Lines  of 
Double  Curvature. 

A  Plane  Curve  is  a  line  moving  always  in  a  plane,  changing  its 
direction  in  accordance  with  some  law.  Examples,  circles, 
ellipses,  etc. 

A  Space  Curve  is  a  curved  line,  no  four  consecutive  points  of 
which  are  contained  in  any  plane.  Example,  the  helix. 

Table  of  Lines 

Straight.     Only  one  kind. 
Circle 


Ellipse          I  -     .    0     A. 

,    ,      )  Come  Sections. 
Hvoerbola 


Plane  Curves  <|  Parabola 

I  Involutes 

>  Gearing  Curves. 
|  Cycloidal  Curves       J 

[  Spirals,  etc. 
f  Helix 
Space  Curves  <}  Most  Intersections  of  Curved  Surfaces 

[  Spherical  Epicycloid,  Spherical  Hypocyloid. 
101 


102  J'KACTICAL  DESCRIPTIVE  GEOMETRY 

116.  Projections  of  Curves. 

A  curve  is  projected  on  H,  V  and  P  in  the  same  manner  as  a 
straight  line;  that  is,  all  the  points  in  the  curve  are  projected  on 
the  plane  of  projection  in  lines  perpendicular  to  the  plane,  and 
the  sum  of  these  projections  of  points  makes  up  the  projection  of 
the  line.  As  these  projectors  are  all  parallel,  the  surface  which 
they  compose  is  called  the  Projecting  Cylinder  of  the  curve. 
(Compare  with  the  projecting  plane  of  a  straight  line.) 

Usually  a  small  number  (six  or  eight)  points  in  space  will  very 
accurately  determine  a  curve.  Its  projections  are  drawn  by 
running  a  smooth  line  through  the  projections  of  the  points  in 
order,  on  both  H  and  V,  and  on  P}  if  necessary. 

Note. — Plane  curves,  in  certain  positions,  will  project  on  one  plane  as  a 
straight  line,  and  the  projecting  cylinder  is  in  that  case  a  projecting  plane. 
However,  most  curves  do  not  lie  so  conveniently,  and  space  curves  never 
project  as  a  straight  line.  Any  curve,  that  is  parallel  or  oblique  to  the 
plane  of  projection,  will  project  as  a  curve  on  that  plane.  Any  plane  curve, 
lying  in  a  plane  perpendicular  to  the  plane  of  projection,  will  project  as  a 
straight  line  on  that  plane. 

117.  Plane  Curves  lying  in  a  Profile  Plane. 

No  curve,  lying  in  a  profile  plane,  can  be  shown  in  its  true 
character  by  the  H-  and  7-projections,  because  they  will  both  be 
straight  lines.  The  P-projection  will  show,  not  only  the  true 
character  of  the  curve,  but  its  true  dimensions.  The  H-  and  V- 
projections  of  a  number  of  points  in  a  curve  will  determine  it, 
but  will  not  show  it. 

In  a  similar  way,  any  curve,  lying  in  a  plane  parallel  to  H,  is 
projected  in  its  true  size  on  H,  and  as  a  straight  line  on  V. 

If  the  curve  lie  in  a  plane  parallel  to  V,  its  F-projection  gives 
its  true  size,  and  its  //-projection  will  be  a  straight  line. 

If  a  curve  lie  in  a  plane  perpendicular  to  H  and  oblique  to  V, 
its  //-projection  will  be  a  straight  line,  and  the  F-projection  will 
be  curved,  and  will  show  the  character  of  the  curve.  The  curve, 
however,  may  be  entirely  changed  by  this  foreshortening;  e.g., 
the  projection  of  an  oblique  circle  is  an  ellipse. 

The  H-  and  F-piercing  points  of  a  curve  are  found  by  the  same 
rule  that  obtains  for  straight  lines;  that  is,  where  the  "^-projection 
crosses  GL,  project  to  the  /^-projection  to  find  the  //-piercing 
point,  etc. 


CURVED  LINES  AND  SURFACES  103 

THEOREMS 

Therein  XXIX. — Any  curve  in  a  plane  perpendicular  to  a 
plane  of  projection  is  projected  thereon  as  a  straight  line. 

Theorem  XXX. — Any  curve  lying  in  a  plane  parallel  to  a  plane 
of  projection  is  projected  thereon  in  its  true  size. 

Theorem  XXXI. — No  space  curve  is  ever  projected  as  a  straight 
line. 

118.  Tangents  and  Normals  to  Curves. 

Definition. — A  tangent  to  a  curve  is  a  line  containing  two 
consecutive  points  of  the  curve. 

The  tangent  may  be  straight  or  curved,  but,  when  curved,  it 
is  always  especially  designated.  When  spoken  of  as  a  tangent, 
the  understanding  is  that  it  is  a  straight  line  tangent. 

119.  The  tangent  (from  its  definition)  must  lie  in  the  plane 
of  the  curve  at  the  point  of  tangency.     Also  (from  its  definition) 
it  must  have  the  same  direction  as  the  curve  at  the  point  of 
tangency.     It   is   especially   important   to   remember   that   the 
tangent  must  lie  in  the  plane  of  the  curve,  as  many  errors  appear 
from  disregarding  this  fact. 

Note. — The  apparent  discrepancy  between  the  foregoing  definition  of  the 
tangent  and  that  given  in  most  Plane  Geometries  amounts  to  nothing. 
They  define  a  tangent  as  a  line  touching  a  curve  at  one  point.  Their  defini- 
tion answers  sufficiently  for  beginners  in  Geometry,  and  for  a  section  of  the 
subject  that  confines  its  operations  to  a  single  plane,  but  the  student  can 
easily  see  how  such  a  definition  would  be  utterly  inadequate  in  Space 
Geometry. 

Note. — A  curve  that  is  tangent  to  a  curve  may  contain  more  than  two 
consecutive  points  of  the  line  to  which  it  is  tangent.  It  brings  about 
no  new  relations  by  reason  of  this  fact,  and  such  a  condition  may  be 
overlooked. 

120.  As  the  tangent  is  a  short  straight  line  in  the  curve  (two 
consecutive   points),    the   curve    may   be    considered    as   being 
composed  of  an  infinite  number  of  infinitesimal  tangents,  inter- 
secting consecutively.     Fig.  69  shows  a  series  of  straight  lines, 
A,  B,  C,  etc.,  intersecting  at  short  finite  intervals.     They  com- 
pose so  nearly  a  curve,  that  it  is  easy  to  imagine  that  minute 
divisions  would  compose  an  actual  curve.     From  the  figure  it 
can  easily  be  seen  how  the  tangent  at  any  point  determines  the 
direction  of  the  curve  at  that  point. 


104 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Theorem  XXXII. — A  tangent  to  a  straight  line  is  the  line 
itself. 

Theorem  XXXIII. — At  any  point  on  a  curve,  one  straight  line 
tangent  and  an  infinite  number  and  variety  of  curved  line  tangents 
may  be  drawn. 

Theorem  XXXIV.— All  of  these  tangents  will  also  be  tangent 
to  each  other. 


FIG.  69. 

121.  Projections  of  the  Tangent. 

The  projections  of  the  tangent  to  a  curve  will  be  tangent  to  the 
projections  of  the  curve  on  the  respective  planes.  The  converse 
is  not  necessarily  true,  because  a  line  lying  in  a  different  plane 
from  that  of  the  curve  may  have  its  projection  tangent  to  that 
of  the  curve  on  one  plane,  but  not  on  any  other.  So  we  have  the 
following  theorem. 

Theorem  XXXV. — If  two  projections  of  a  line  and  a  curve  are 
tangent,  the  line  and  curve  are  tangent. 

122.  To  Draw  a  Tangent  to  a  Curve. 

There  are  exact  and  approximate  methods.  In  the  case  of  the 
circle,  the  tangent  is  always  perpendicular  to  the  radius  at  the 
point  of  contact.  In  the  case  of  other  conic  sections,  the  exact 
method  will  be  taken  up  in  connection  with  the  study  of  those 
curves.  For  ordinary  drafting  work  a  sufficiently  accurate 
tangent  may  be  drawn  by  placing  a  straight  edge  against  the 
curve,  and  laying  off  the  tangent  with  the  eye.  A  simple  approxi- 
mation, that  can  be  used  in  nearly  all  cases  of  curves,  and  is  exact 


CURVED  LINES  AND  SURFACES  105 

for  the  circle,  is  this:  Let  it  be  required  to  draw  the  tangent  at 
A  (Fig.  70)  on  the  curve  shown.  Set  the  dividers  at  some  small 
arbitrary  span,  and  lay  off  B  and  C  on  the  curve  at  equal  dis- 
tances from  A.  Connect  B  and  C  with  a  straight  line.  Through 
A  draw  AD  (the  tangent)  parallel  to  BC. 


FIG.  70. — Drawing  a  tangent. 

This  method  will  not  do  for  curves  changing  their  radii  rapidly. 
The  eye  will  do  close  work  in  sharply  curving  lines.  Numerous 
close  approximations  may  be  found  in  other  text-books,  and  an 
especially  good -one  is  given  in  MacCord's  "  Elements  of  Descrip- 
tive Geometry,"  page  93,  but  its  practical  value  is  small. 

123.  Normals. — A  normal  to  a  curve  at  any  point  is  a  perpen- 
dicular to  the  tangent  at  the  point  of  tangency.  While  any 
perpendicular  is  normal  to  the  curve,  it  is  customary  to  regard 
the  normal  as  the  perpendicular  to  the  tangent  in  the  plane  of 
the  curve. 


FIG.  71. — The  rectification  of  a  curved  line. 

In  the  case  of  a  space  curve,  the  normal  may  be  any  line  per- 
pendicular to  the  tangent  at  the  point  of  tangency. 

124.  The  Rectification  of  Curves. 

To  rectify  a  curve  means  to  straighten  it  out;  that  is,  to  draw 
a  tangent  at  any  point  equal  in  length  to  any  given  portion  of 
the  curve. 

First. — To  rectify  a  portion  of  any  curve. 

Let  the  curve  be  AB,  Fig.  71.  Set  the  dividers  conveniently 
small,  so  as  to  lay  off  an  arc,  whose  chord  is  approximately  the 
length  of  the  arc.  Step  off  with  the  dividers  the  points  1,  2,  3,  4, 


106 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


etc.,  from  B  toward  A,  where  the  tangent  is  drawn.  Lay  off 
with  a  different  pair  of  dividers  the  last  step,  5A  (which  is  not 
likely  to  be  the  same  as  the  others),  on  the  tangent  A5'.  Then, 
with  the  original  dividers,  step  off  from  that  point  the  points 
4',  3',  2',  T  and  B'.  This  rectification  is  fairly  good,  if  care  is 
taken. 

Second. — To  rectify  a  small  arc  of  a  circle. 

If  the  arc  be  less  than  60°,  a  convenient  method  is  shown  in 
Fig.  72.  Let  AB  be  the  arc,  and  BD  (of  indefinite  length)  be 
the  tangent.  Extend  the  chord  AB  through  B  one-half  its 
length  to  C.  Using  C  as  a  center,  and  CA  as  a  radius,  strike 
the  arc  AD.  BD  will  then  equal  (approximately)  the  arc  BA. 

Third. — To  rectify  a  larger  arc. 

Divide  the  large  arc  into  smaller  ones,  rectify  each,  and  add 
them  together. 


FIG.  72.— The  rectification  of 
circular  arc. 


FIG.  73.— The  rectification  of 
quadrant. 


Fourth. — To  rectify  a  quadrant  of  a  circle. 

Let  AB  (Fig.  73)  be  the  given  quadrant.  Draw  the  tangent 
at  B,  and  from  A  draw  a  line  AC,  which  makes  60°  with  CB. 
BC  is  then  a  very  close  approximation  to  AB.  This  is  undoubt- 
edly the  closest  approximation  for  such  an  arc.  Let  the  student 

prove   by   trigonometry  that   it   is   numerically   equal   to  •'"-  H. 

Although  there  is  no  geometrical  proof  for  this,  your  calculation 

will  show  that  it  is  closer  than  the  accuracy  of  your  instruments. 

A  convenient  method  of  rectifying  a  small  arc  comes  with 


CURVED  LINES  AND  SURFACES  107 

this  method.     To  rectify  BD,  Fig.  73,  draw  the  line  ED  through 
to  the  tangent  at  F.     The  line  BF  equals  BD. 

Fifth.—  To  rectify  an  arc  of  a  circle  when  its  angle  is  known. 

Here  the  numerical  method  is  best.     The  arc  AB   (angle  =  0) 


125.  To  lay  off  any  length  of  line,  straight  or  curved,  on  any 
curve. 

Regard  the  line  AB'  (Fig.  71)  as  a  curved  line.  Lay  off  AB 
on  the  curve  in  exactly  the  same  manner  that  it  was  laid  off  on 
the  straight  line. 

This  is  often  a  necessary  operation  in  practical  drafting; 
example,  in  drawing  cycloids  for  gear  outlines. 

126.  Conic  Sections. 
Definitions. 

Circle.  —  A  circle  is  a  plane  curve  generated  by  a  point  moving 
so  as  to  be  at  a  constant  distance  from  another  point. 

Ellipse.  —  An  ellipse  is  a  plane  curve  generated  by  a  point 
moving  so  that  the  sum  of  its  distances  from  two  fixed  points  in 
its  plane  is  constant. 

Hyperbola.  —  A  hyperbola  is  a  plane  curve  generated  by  a  point 
moving  so  that  the  difference  between  its  distances  from  two 
fixed  points  in  its  plane  is  constant. 

Parabola.  —  A  parabola  is  a  plane  curve  generated  by  a  point 
moving  so  that  its  distances  from  a  line  and  a  point  in  the  plane 
are  always  equal. 

The  four  foregoing  curves  are  called  "Conic  Sections"  from 
the  fact  that  they  are  the  possible  curves  that  can  be  cut  from  a 
right  circular  cone  by  a  plane.  A  plane  can  also  cut  a  straight 
line  from  a  cone.  The  necessary  conditions  for  cutting  these 
lines  from  a  cone  are: 

Straight  Line.  —  Plane  through  the  apex. 

Circle.  —  Plane  perpendicular  to  the  axis. 

Ellipse.  —  Plane,  oblique  to  the  axis,  cutting  all  the  elements 
on  one  side  of  the  apex. 

Hyperbola.  —  Plane,  parallel  to  the  axis,  or  a  plane  oblique  to 
the  axis  at  a  smaller  angle  than  that  made  by  the  elements  of  the 
cone. 

Parabola.  —  Plane  parallel  to  one  element  of  the  cone. 

127.  Gearing  Curves. 

Cycloid.  —  A  cycloid  is  a  plane  curve  generated  by  a  point  on 


108  PRACTICAL  DESCRIPTIVE  GEOMETRY 

the  circumference  of  a  circle,  as  the  circle  rolls  on  a  straight 
line. 

Epicycloid. — An  epicycloid  is  a  plane  curve  generated  by  a 
point  on  the  circumference  of  a  circle,  as  it  rolls  on  another  circle. 

Hypocycloid. — A  hypocycloid  is  a  plane  curve  generated  by  a 
point  on  the  circumference  of  a  circle  as  it  rolls  on  the  inside  of 
another  circle. 

Involute. — An  involute  is  a  plane  curve  generated  by  a  point 
on  a  tangent,  as  the  tangent  rolls  on  a  plane  curve  or  polygon. 
The  involute  of  a  circle  is  the  only  one  of  any  considerable 
importance,  and  is  the  curve  used  in  gearing. 

The  foregoing  are  called  "  gearing  curves,"  because  they  are 
mainly  used  as  the  outlines  of  teeth  in  gear  wheels.  The  outlines 
of  bevel  gears  are  (theoretically)  spherical  cycloids,  but  their 
consideration  is  negligible  in  this  work,  and  practically  so  in 
actual  drafting  work. 

SPACE  CURVES 

128.  Helix. — A  helix  is  a  curve  generated  by  a  point  moving 
about  a  straight  line  at  a  constant  distance,  and  in  the  direction 
of  the  line  at  a  constant  velocity. 

Note. — This  line  is  the  line  of  the  screw  thread,  twist  drill,  screw  con- 
veyor, worm  gear,  coiled  spring,  and  many  other  important  practical 
devices.  It  is  a  curve  on  the  surface  of  a  cylinder. 

There  is  also  a  conical  helix,  that  sometimes  appears  in  practical 
work.  It  is  usually  drawn  so  that  it  makes  one  complete  revolu- 
tion about  the  axis  in  travelling  from  the  base  to  the  apex,  ascend- 
ing the  surface  at  a  uniform  rate  of  motion.  The  plan  of  the 
conical  helix  is  the  Spiral  of  Archimedes.  By  drawing  the  plan 
first  and  projecting  the  points  of  the  curve  as  it  intersects  the 
various  elements  of  the  cone,  the  elevation  is  easily  drawn.  It 
is  rather  commonly  used  in  sheet  metal  work  and  in  spiral  springs. 

There  is  no  need  of  defining  or  considering  any  other  of  the 
space  curves. 

GENERAL  DEFINITIONS 

129.  Axis. — The  line  about  which  a  point  revolves,  as  in  the 
case  of  the  helix.     It  is  also  a  center  line  for  some  curves,  as  the 
ellipse,  parabola,  and  hyperbola. 


CURVED  LINES  AND  SURFACES  109 

Minor  Axis  and  Major  Axis. — In  the  ellipse  there  are  two 
dimensions,  sometimes  called  the  long  and  short  diameters,  but 
more  generally  the  major  and  minor  axes.  They  are  respectively 
the  longest  and  shortest  possible  straight  lines  running  through 
the  center  from  one  point  in  the  curve  to  another.  The  constant 
sum,  spoken  of  in  the  definition,  is  equal  to  the  length  of  the 
major  axis. 

Focus. — The  two  fixed  points,  mentioned  in  the  definitions 
of  the  ellipse  and  hyperbola,  and  the  single  one,  mentioned  in 
the  definition  of  the  parabola. 

Vertex. — The  nearest  point  in  the  curve  to  the  focus.  It  has 
the  smallest  radius  of  curvature  of  any  point  in  the  curve,  and 
might  be  described  as  the  center  of  symmetry  of  the  curve.  The 
circle  has  no  vertex. 

The  constant  difference,  mentioned  in  the  definition  of  the 
hyperbola,  is  the  distance  between  the  two  vertices  of  the  curve. 

Directrix. — The  straight  line  mentioned  in  the  definitions  of 
the  parabola  and  the  cycloid. 

Generating  Circle. — The  rolling  circles  in  the  definitions  of  the 
cycloid,  epicycloid  and  hypocycloid. 

Pitch  Circle,  or  Directing  Circle. — The  circle  on  which  the 
generating  circle  rolls. 

Evolute. — The  curve  on  which  the  tangent  generating  the 
involute  rolls. 

Note. — The  tangent  to  the  evolute  is  normal  to  the  involute. 

Pitch. — The  pitch  of  a  helix  is  the  distance  that  it  travels  in 
the  direction  of  the  axis  in  making  one  complete  revolution. 

PROBLEMS  RELATING  TO  CURVED  LINES 

130.  Problem  44.— To  draw  an  Ellipse,  having  given  its  foci 
and  "constant  sum." 

Construction. — 1.  Draw  the  line  AB  (Fig.  74)  and  from  a 
center,  O,  lay  off  F  and  F'  at  equal  distances  from  O,  making 
FF'  equal  to  the  given  distance  between  the  foci. 

2.  On  the  same  line  lay  off  A  and  B  at  equal  distances  from  O, 
making  AB  equal  to  the  given  constant  sum. 

3.  Spread  the  compasses  equal  to  AO,  and  strike  arcs  from 
F  and  F'  as  centers.     The  intersections,  C  and  D,  of  these  arcs 
are  the  ends  of  the  minor  axis. 


110 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


4.  Set  the  compasses   at  arbitrary  radii,  Al;  A2,   etc.,   and 
strike  arcs  with  them  from  F  and  F'  as  centers. 

5.  Set  the  compasses  at  complementary  radii,  Bl,  B2,  etc., 
and  strike  arcs  from  F  and  F'  as  centers. 

6.  Through  A,  B,  C,  D,  and  the  intersections  of  complementary 
radii,  draw  a  smooth  line.     This  will  be  the  required  ellipse. 

Proof. — The  sum  of  the  distances  of  any  of  the  points,  thus 
found,  to  the  foci  is  always  equal  to  AB. 

"Note. — There  are  numerous  ways  of  drawing  an  ellipse,  exact  and  ap- 
proximate, which  are  found  in  any  text-book  on  Mechanical  Drawing. 
Probably  the  quickest  and  most  convenient  is  the  "trammel  method,"  and 
next  easiest  is  the  "concentric  circles  method."  Any  method  whereby  the 
compasses  are  used  to  describe  the  curve  is  necessarily  inexact,  because  no 
arc  of  a  circle  is  any  arc  of  an  ellipse.  However,  in  practical  work,  it  is 
often  expedient  and  desirable  to  use  such  ellipses,  especially  in  Isometric 
Projection. 


FIG.  74.— The  ellipse. 

131.  Uses  for  the  Ellipse. — In  ornamental  work  the  ellipse  is 
extensively  used,  in  its  entirety,  and  in  portions.     The  ellipse 
is  the  projection  of  any  circle  in  an  oblique  plane,  and  its  major 
axis  is  always  equal  to  the  diameter  of  the  circle.     In  machine 
design,  the  ellipse  is  used  in  certain  " quick  return"  motions. 

Limits. — When  the  distance  between  the  foci  is  zero,  the 
ellipse  becomes  a  circle.  When  the  distance  between  the  foci 
equals  the  "  constant  sum,"  the  ellipse  becomes  a  straight  line. 

132.  Problem  45. — To  draw  the  tangent  to  an  Ellipse  at  a 
given  point  on  the  curve. 


CURVED  LINES  AND  SURFACES 


111 


Construction. — Let  M  (Fig.  74)  be  the  point. 

1.  Draw  MF  and  MF',  the  focal  lines. 

2.  Extend  MF'  beyond  M  to  F". 

3.  Bisect  the  exterior  focal  angle  FMF". 
The  bisector  MX  will  be  the  tangent  at  M. 

133.  Problem  46. — To  draw  the  tangent  to  an  Ellipse  from  a 
point  outside. 

Let  N  (Fig.  74)  be  the  point. 

1.  Draw  a  circle  with  N  as  center  and  NF'  as  radius. 

2.  Intersect  the  circle   in  points  P  and  Q  by  an   arc  whose 
radius  is  equal  to  AB,  and  whose  center  is  at  F. 

3.  Connect  either  P  or  Q  with  F. 

4.  Where  PF  or  QF  intersect  the  ellipse  at  R  or  S,  will  be  the 
points  to  which  tangents  can  be  drawn  from  N. 

134.  If  the  axes  of  an  ellipse  are  given,  the  foci  may  be  deter- 
mined by  striking  an  arc  from  C  or  D  as  center,  with  a  radius 
equal  to  AO.     Where  this  arc  intersects  AB  will  be  the  foci. 


FIG.  75.— The  hyperbola. 

135.  Problem  47.— To  draw  a  Hyperbola,  having  given  its  foci 
and  the  "constant  difference." 

Let  F  and  F'  (Fig.  75)  be  the  foci,  and  the  "constant  difference" 
be  equal  to  the  distance  VV. 

Construction. — 1.  Lay  off  F  and  F'  and  V  and  V  on  the  same 
line,  at  their  proper  distances,  all  symmetrically  placed  about  a 
center,  O.  (V  and  V'  will  be  the  vertices.) 

2.  Spread  the  compass  at  any  arbitrary  spans,  as  VI,  V2,  etc., 
and  strike  arcs  with  these  radii  with  F  and  F'  as  centers. 


112  PRACTICAL  DESCRIPTIVE  GEOMETRY 

3.  Using  the  complementary  radii,  VI,  V2,  etc.,  strike  arcs 
with  F  and  F'  as  centers,  intersecting  those  already  drawn. 

4.  Draw  a  smooth  curve  through  the  proper  intersections  and 
V  and  V.     This  will  be  the  required  hyperbola. 

Note. — The  hyperbola  is  in  reality  two  curves,  each  having  a  vertex,  and 
each  unlimited  in  extent.  The  hyperbola  varies  in  accordance  with  the 
ratio  between  FF'  and  VV.  The  most  important  use  of  the  hyperbola  is 
that  it  is  the  generating  line  of  the  pitch  surface  of  "skew"  gears. 

136.  Problem  48. — To  draw  the  tangent  to  a  Hyperbola  from 
any  point  of  the  curve. 

Construction.— Let  P  be  the  point. 

1.  Draw  PF  and  PF',  the  focal  lines. 

2.  Bisect  the  angle  FPF'. 

The  bisector  will  be  the  tangent. 

Note. — The  construction  is  not  shown  in  the  figure.  Let  the  student 
make  it  for  himself. 

137.  Problem  49. — To  draw  a  tangent  to  a  Hyperbola  from 
any  point  outside  the  curve. 

Let  N  (Fig.  75)  be  the  given  point. 

Construction. — 1.  With  N  as  a'  center,  draw  a  circle  passing 
through  F*,  the  nearest  focus. 

2.  With  Fj  the  farther  focus,  as  the  center,  and  VV  as  the 
length  of  the  radius,  strike  an  arc,  cutting  the  first  circle  in  the 
points  X  and  Y. 

3.  Draw  XF^and  YF'  extending  them  until  they  intersect  the 
hyperbola  in  the  points  P  and  Q. 

4.  NP  and  NQ  will  be  the  required  tangents. 

138.  Problem   50. — To   draw   a  Parabola,   having   given  the 
focus  and  the  directrix. 

Construction. — Let  XD  (Fig.  76)  be  the  directrix  and  F  the 
focus. 

1.  Draw  a  line  CG  through  F  perpendicular  to  XD.     This  is 
the  axis  of  the  curve. 

2.  Bisect  the  distance  from  F  to  XD  at  E.     E  will  be  the 
vertex. 

3.  Draw  a  number  of  lines  at  points  1,  2,  3,  etc.,  parallel  to 
XD,  at  arbitrary  distances. 

4.  Locate  two  points  on  each  of  the  lines,  equally  distant  from 
F  and  XD. 

5.  Draw  a  smooth  line  through  the  points  thus  found,  and  it 
will  be  the  required  parabola. 


CURVED  LINES  AND  SURFACES 


113 


139.  Problem  51. — To  draw  the  Tangent  to  a  Parabola  at  any 
point  on  the  surface. 

Construction. — Let  M  (Fig.  76)  be  the  given  point. 

1.  Draw  MX,  perpendicular  to  XD. 

2.  Draw  the  focal  line  MF. 

3.  Bisect  the  angle  FMX.     The  bisector  is  the  tangent. 

Note. — Extend  the  line  XM  in  the  direction  MY,  and  draw  the  normal 
MO.  Notice  that  MY  is  parallel  to  the  axis,  and  that  MO  is  the  bisector 
of  the  angle  FMY,  making  OMF  equal  to  OMY.  This  fact  is  taken  advan- 
tage of  in  designing  the  reflector  for  the  searchlight.  M  is  a  random  point 
on  the  parabola,  and  shows  that  any  ray  from  F,  striking  the  parabola  at 
M  (any  point)  is  reflected  in  a  line  parallel  to  the  axis,  as  MY.  This  con- 
centrates all  the  reflected  rays  into  a  cylinder  of  light,  preventing  diffusion. 


FIG    76. — The  parabola. 

140.  Problem  52. — To  draw  a  Tangent  to  a  Parabola  from  a 
point  outside. 

Construction. — Let  N  (Fig.  76)  be  the  given  point. 

1.  With  N  as  a  center,  draw  a  circle  through  F. 

2.  From   its   intersections,   C   and   D,   with   DX,   draw    the 
perpendiculars,  CE  and  DK. 

3.  E  and  K,  the  intersections  with  the  curve,  are  the  points  to 
which  tangents  can  be  drawn. 

141.  Problem  53. — To  draw  a  Cycloid,  having  given  the  direc- 
trix and  generating  circle. 


114  PRACTICAL  DESCRIPTIVE  GEOMETRY 

Construction. — Let  AB  be  the  directrix  and  O  be  the  center 
of  the  generating  circle,  which  is  to  be  drawn  tangent  to  the 
directrix,  of  a  radius  R. 

1.  Lay  off  on  AB;  a  distance  equal   to  2;rR,  and  divide  the 
circle  and  AB  into  any  convenient  number  of  equal  parts,  12, 
16  or  24. 

2.  Draw  the  path  of  the  center  of  the  cirele,  parallel  to  AB, 
at  a  distance  equal  to  R. 

3.  Erect  perpendiculars  to  the  line  of  centers  at  each  of  the 
divisions  of  AB.     The  intersections  will  be  successive  positions 
of  the  center  of  the  generating  circle. 


A  P  B 

FIG.  77.— The  cycloid. 

4.  From  the  divisions,  1,  2,  3,  etc.,  of  the  generating  circle, 
run  projectors  parallel  to  AB. 

5.  From    each    of   the    successive   positions    of   the    center, 
I',    2',    3',   etc.,   describe    a^cs    intersecting    the  corresponding 
projectors. 

6.  Through  these  intersections  draw  a  smooth  curve,  which 
will  be  the  required  Cycloid. 

142.  Problem  54. — To  draw  the  tangent  to  the  Cycloid  at  any 
point  on  the  curve. 

Construction. — Let  M  (Fig.  77)  be  the  given  point,  and  P  be 
the  point  of  tangency  with  AB  of  the  generating  circle,  when  M 
is  its  intersection  with  the  curve. 

1.  Draw  MP.     This  line  is  the  normal  at  M. 

2.  Draw  the  tangent  perpendicular  to  MP. 

Note. — It  is  owing  to  the  fact  that  the  normal  to  the  cycloid  at  any  point 
passes  through  the  point  P,  where  the  generating  circle  and  directrix  are 
tangent,  that  the  cycloid  is  a  correct  curve  for  use  in  gear  outlines.  It  is 
this  property  that  makes  it  possible  to  maintain  a  constant  angular  veloc- 
ity ratio  in  gears,  as  will  be  seen  when  the  study  of  Kinematics  is 
taken  up. 


CURVED  LINES  AND  SURFACES  115 

143.  Problem  55. — To  draw  an  Epicycloid  and  a  Hypocycloid. 

The  only  difference  between  drawing  these  curves  and  the 
cycloid,  is  that  the  directrix  now  is  a  circle  instead  of  a  straight 
line.  Fig.  78  shows  both  curves.  AB  is  an  arc  of  the  pitch 
circle,  as  the  directrix  is  usually  called  in  Machine  Design,  and 
the  center  of  the  generating  circle  of  the  Epicycloid  is  at  C,  and 
its  path  is  the  arc  CD  (center  at  O) .  The  center  of  the  Hypo- 
cycloidal  circle  is  at  E,  and  its  path  is  the  arc  EF,  center  at  0. 


FIG.  78. — The  epicycloid  and  hypocycloid. 

Construction. — -1.   Divide  the  generating  circles  into  any  con- 
venient number  of  equal  parts. 

2.  Step  off  these  arcs  on  AB  (Art.  125). 

3.  From  the  points  1,  2,  3,  etc.,  draw  arcs  with  their  centers 
at  O. 

4.  Through  the  divisions  on  AB  draw  radial  lines  intersecting 
CD  and  E^  in  the  points  occupied  by  the  centers  as  the  circles 
roll  on  AB. 

5.  From  these  center  points  strike  arcs  with  radii  equal  to 
those  of  the  generating  circles,  intersecting  the  arcs  of  opera- 
tion (3). 

6.  Through  the  respective  intersections  draw  the  curves. 
The  tangents  are  drawn  precisely  as  the  tangents  to  the  cycloid 

are  drawn;  that  is,  the  normal  at  any  point  passes  through  the 
point  of  tangoncy  of  the  generating  circle  and  the  pitch  circle. 


116 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


144.  Problem  56. — To  draw  an  Involute  to  a  Circle. 

Let  AB  (Fig.  59)  be  the  directing  circle,  or  Evolute,  with  its 
center  at  O. 

Construction. — 1.  Divide  any  arc  of  AB,  as,  for  example,  the 
semicircle,  into  any  number  of  equal  parts. 

2.  At  the  points  1,  2,  3,  etc.,  draw  tangents  in  the. direction  of 
the  origin,  A. 

3.  Rectify  each  arc  from  the  origin  and  lay  off  the  distance  on 
the  tangent. 

4.  Through  the  extremities  of  the  tangents  draw  the  curve. 


FIG.  79.— The  involute. 

Note  1. — This  curve  is  that  described  by  the  end  of  a  taut  cord  as  it  is 
unwound  from  a  post.  It  is  not  a  closed  curve,  but  a  spiral  that  keeps 
growing  in  expanse  with  every  revolution. 

Note  2. — An  involute  can  be  described  in  the  same  way,  using  any  curve 
or  polygon  as  the  evolute. 

Note  3. — The  Volute,  used  in  architecture,  is  a  spiral,  the  involute  of  an 
enlarging  square. 

145.  Problem  57. — To  draw  the  Tangent  to  an  Involute  at  any 
point  on  the  curve. 

The  tangents  to  the  evolute  are  all  normals  to  the  involute 
(Art.  129) ,  hence  all  that  is  necessary  is  to  draw  the  perpendicular 
to  the  normal  at  the  given  point.  Thus  at  the  point  C  (Fig.  79) 
draw  CD  perpendicular  to  C8,  the  normal. 

Note. — The  fact  that  the  normal  to  the  involute  is  tangent  to  the  evo- 
lute makes  the  curve  useful  in  gearing  in  the  same  way  as  in  the  case  of 
cycloidal  curves. 

146.  Problem  58. — To  draw  the  projections  of  a  Helix,  having 
given  its  Pitch  and  Diameter. 

Let  OP  (Fig.  80)  be  the  axis,  o'12  the  pitch,  and  D  the  diameter. 


CURVED  LINES  AND  SURFACES 


117 


Construction. — 1.   Draw  the  circle  and  divide  it  into  any  num- 
ber of  equal  parts. 

2.  Divide  the  pitch  into  the  same  number  of  equal  parts. 

3.  Draw  perpendiculars  to  the  axis  from  each  of  the  divisions 
made  in  operation  (2). 

4.  Draw  projectors   from  each  of   the  divisions  of  the  circle, 
intersecting  the  corresponding  perpendiculars  from  the  axis. 

5.  Draw  the  curve  through  these  intersections. 


FIG.  80.— The  helix. 

147.  The  Angle  of  the  Helix. 

From  the  definition  (Art.  128),  we  see  that  the  generating 
point  on  the  helix  ascends  a  given  distance  in  traversing  a  given 
arc,  and  that  the  rate  of  ascent  is  constant.  This  means  that 
any  element  of  the  curve  makes  a  certain  angle  with  the  horizontal 
(the  axis  of  the  helix  being  taken  in  all  these  problems  as  vertical) , 
and  that  all  the  elements  in  any  one  helix  make  the  same  angle 
with  the  horizontal.  Thus,  in  Fig.  80  (a),  if  AB  is  the  length  of 
an  elemental  arc  of  the  circle,  and  BC  is  the  length  of  an  elemental 
division  of  the  pitch,  the  hypothenuse  of  the  right  triangle  ABC 
will  yield  the  inclination  and  the  length  of  an  element  of  the 


118 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


helix.     Therefore,  if  a  right  triangle,  having  its  base  angle  equal 
to  a,  be  wrapped  about  a  cylinder,  the  hypothenuse  will  become 


the  helix. 


pitch 

From  this  it  is  evident  that  tan.  a  = . — 

7i  Diam. 


148.  The  Tangent  to  the  Helix. 

If  the  right  triangle,  just  mentioned,  be  unwrapped  from  the 
cylinder,  its  hypothenuse  becomes  the  tangent  at  the  point  where 
the  unwrapping  ceases. 

This  is  so  because  the  tangent  must  make  the  same  angle 
with  the  horizontal  that  the  helix  does,  because  it  contains  two 
consecutive  points  (one  element)  of  the  curve. 


FIG.  81. — The  generation  of  a  helix. 

Therefore,  the  tangent  becomes  the  helix  if  it  is  rolled  on  the 
cylinder,  and  the  length  of  the  helix  from  its  origin  to  the  point  of 
tangency  is  equal  to  the  length  of  the  tangent  from  the  plane  of 
the  origin  to  the  point  of  tangency.  Referring  to  the  picture, 
Fig.  81,  will  show  that  AO  =  BO,  and  further  that  AP  =  BP. 
AP  is  the  //-projection  of  the  tangent,  and  BP  is  the  //-pro- 
jection of  the  helix.  From  this  the  rule  is  derived:  The  //- 
projection  of  the  Helix  is  of  the  same  length  as  the  //-projection  of 
the  Tangent  from  the  plane  of  the  origin  to  the  point  of  tangency. 

149.  Problem  59. — To  draw  the  Tangent  to  the  Helix  at  any 
point  on  the  curve. 

Construction. — Let  B  (Fig.  80)  be  the  given  point.  1.  Draw 
the  tangent  to  the  circle  at  b. 

2.  Rectify  the  arc  b!2,  and  lay  it  off  on  the  tangent  ab. 

3.  Project  a  to  a'  on  GL. 

4.  a'b'  will  be  the  F-projection,  and  ab  the  //-projection  of 
the  tangent  required. 


CURVED  LINES  AND  SURFACES  1  19 


150.  EXERCISES  IN  CURVED  LINES 

Note. — No  ground  line  is  necessary  in  drawing  any  of  the  curves  but  the 
helix. 

518.  Draw  an  Ellipse,  AB  =  5  in.,  FF'  =  3  in.     Draw  a  tangent  at  a  point  on 
the  curve  2  in.  from  F.     Draw  two  tangents  from  a  point  4  in.  from  F 
and  2  in.  from  F'. 

519.  Draw  an  Ellipse,  AB=4£  in.,  and  CD  =  3  in.     Locate  F  and  F',  and 
draw  a  tangent  at  a  point  on  the  curve  1  in.  from  A.     Draw  two 
tangents  from  E,  which  is  3  in.  from  F'  and  2  in.  from  F. 

520.  A  and  B  are  two  tangent  ellipses  of  equal  size.     The  foci  of  A  are 
C(l,  0-2)D(3£,  0-2),   and   the   more   distant   focus   of    B   is   at   E 
(5,  0  +  1).     Draw  the  two  ellipses. 

521.  Draw  a  Hyperbola.     FF'=3  in.  and  the  "constant  difference"  =2  in. 
Draw  the  tangent  at  a  point  on  the  curve  2  in.  from  F.     Draw  a 
tangent  from  a  point  2  in.  from  F'  and  2^  in.  from  F. 

522.  Draw  a  Parabola  with  the  distance  between  focus  and  directrix  1  in. 
Draw  tangents  from  assumed  points  on  and  outside  the  curve. 

523.  Draw  a  Parabola  whose  distance  from  focus  to  vertex  is  f  in.     Draw 
tangents  from  assumed  points  on  and  outside  the  curve. 

524.  Draw  a  Cycloid  with  a  1^-in.  generating  circle.     Draw  a  tangent  at  the 
point  on  the  curve  at  -fs  of  the  revolution. 

525.  Draw  an  Epicycloid  and  a  Hypocycloid  on  a  pitch    circle  of    6-in. 
radius,   with   1^-in.   generating   circles.     Draw  the   normals   to   each 
curve  at  T5ff  of  a  revolution  from  the  origin. 

526.  Draw  an  Involute  of  a  1-in.  circle. 

527.  Draw  an  Involute  of  a  f-in.  square. 

528.  Draw  an  Involute  of  a  hexagon  of  J-in.  sides. 

529.  Draw  an  Involute  of  a  f-in.  equilateral  triangle. 

530.  Draw  a  Helix  of  2-in.  diameter,  3-in.  pitch.     Draw  its  tangent  at  60° 
from  the  origin.     Also  one  at  90°. 


SURFACES 

151.  Definitions. 

A  line,  straight  or  curved,  moving  in  accordance  with  some 
law,  generates  a  Surface.  If  the  generating  line  be  straight,  and 
moves  along  two  parallel  or  intersecting  straight  lines,  the 
surface  generated  will  be  a  Plane.  There  is  but  one  kind  of 
plane,  all  other  kinds  of  surfaces  are  curved. 

152.  Curved  Surfaces  are  divided  into  three  classes,  Single- 
Curved,  Double -Curved  and  Warped  surfaces. 

A  Single -Curved  Surface  is  generated  by  a  straight  line  follow- 
ing a  curved  director  in  such  a  way  that,  any  two  consecutive 
positions  of  the  line  shall  lie  in  a  plane. 


120 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


A  Warped  Surface  is  generated  by  a  straight  line  moving  so 
that  no  two  of  its  consecutive  positions  shall  be  in  the  same  plane. 

Note. — Single-curved  and  warped  surfaces  are  sometimes  called  Ruled 
surfaces,  because  they  are  made  up  of  straight  lines.  Through  every  point 
of  a  ruled  surface  at  least  one  straight  line  can  be  drawn  in  the  surface. 

A  Double -Curved  Surface  is  generated  by  a  curve  moving 
along  a  curved  line  director. 

Note. — There  can  be  no  straight  lines  in  any  double-curved  surface. 

Generatrix. — The  line  generating  a  surface. 

Element. — Any  position  of  the  generatrix. 

Consecutive  Elements. — Two  successive  positions  of  the 
generatrix,  having  no  assignable  distance  between  them. 

A  Surface  of  Revolution  is  a  surface  generated  by  the  revolu- 
tion of  any  line,  straight  or  curved,  about  a  straight  line  axis. 
Surfaces  of  revolution  may  be  single-curved,  double-curved,  or 
warped. 

153.  Classification  of  Surfaces. 

Single -Curved  Surfaces  are  of  three  varieties: 

Cones,  all  the  elements  intersect  in  a  point. 

Cylinders,  all  the  elements  are  parallel. 

Convolutes,  consecutive  elements  intersect  two  and  two,  no 
three  elements  having  a  common  point. 

Double  Curved  Surfaces  are  mostly  surfaces  of  revolution, 
spheres,  ellipsoids,  paraboloids,  tori,  and  combination  surfaces. 
They  are  sometimes  convex,  sometimes  concave,  and  sometimes 
both.  The  annular  torus  has  both  in  continuous  surface. 

Warped  Surfaces  are  numerous  in  their  variety,  but  the  most 
important  are  the  helicoid  and  hyperboloid  of  revolution  of  one 
nappe. 

154.  The  following  table  will  give  a  convenient  summary  of 
the  surfaces. 

f  Cone — all  elements  intersecting 

Cylinder — all  elements  parallel 

Convolute — consecutive  elements  intersect- 
I       ing. 

Helicoid 

Hyperboloid  of  Revolution  [•  Ruled  Surfaces. 

Hyperbolic  Paraboloid 

Conoid 

Cylindroid 

Warped  Cone 

Cow's  Horn. 


Single- 
Curved 


Warped 


CURVED  LINES  AND  SURFACES  121 

Sphere 

Oblate  Spheroid 
Double-  j  Prolate  Spheroid 
Curved  j  Paraboloid  (both  varieties) 

|  Hyperboloid  of  Revolution  of  Two  Nappes 
[  Torus. 

Surfaces  of    [  Right  Circular  Cone,  Right  Circular  Cylinder 
Revolution   \  Hyperboloid  of  Revolution  of  One  Nappe 
[  Sphere,  Spheroids,  O voids,  etc. 

155.  Tangent  Planes  and  Developments. 

The  most  important  problems  in  surfaces  are  to  be  found  in 
drawing  tangent  planes  to  surfaces,  the  intersections  of  surfaces, 
and  developing  the  surfaces. 

A  Tangent  Plane  to  a  surface  at  any  point  is  a  plane  composed 
of  all  the  tangent  lines  to  all  the  lines  of  a  surface  passing  through 
that  point.  As  two  intersecting  lines  determine  a  plane,  the 
general  rule  for  tangent  planes  is  this:  Any  two  tangent  lines  at 
any  point  on  any  surface  determine  the  tangent  plane  at  that 
point. 

Any  plane,  passed  through  the  point  of  tangency  so  as  to  in- 
tersect a  surface  and  its  tangent  plane,  will  cut  from  them  a  line 
and  its  tangent. 

In  ruled  surfaces,  one  and  sometimes  two  straight  lines  of 
the  surface  can  be  drawn  through  each  point  in  the  surface, 
therefore  this  line  (element,  as  it  always  is)  can  be  used  for  one 
of  the  tangents  in  the  foregoing  rule.  This  is  true,  because  a 
straight  line  is  its  own  tangent.  On  this  account,  the  rule  wrill 
read:  uln  ruled  surfaces,  one  tangent  and  the  element  at  any 
point  will  determine  the  tangent  plane  at  that  point." 

Normal. — The  normal  line  at  any  point  of  a  curved  surface  is 
the  perpendicular  to  the  tangent  plane  at  that  point.  The 
normal  plane  is  any  plane  containing  the  normal  line. 

Development. — By  Development  is  meant  the  rolling  out  of  a 
surface  into  a  plane  figure,  which  can  be  rolled,  or  folded,  back 
to  make  the  original  surface.  Only  single-curved  surfaces  are 
strictly  developable,  because  they  only  have  consecutive  elements 
lying  in  planes.  Approximate  developments  can  be  made  of 
other  surfaces,  as  will  appear  when  the  surfaces  are  taken  up 
separately. 


122  PRACTICAL  DESCRIPTIVE  GEOMETRY 

PROBLEMS  RELATING  TO  SINGLE-CURVED  SURFACES 

156.  The  Cone. 

Definitions. — A  Cone  is  generated  by  a  straight  line  moving  so 
as  to  constantly  touch  a  curved  line  directrix,  while  one  point 
in  the  generatrix  remains  stationary. 

Apex  or  Vertex,  the  stationary  point  on  the  generatrix. 

Base,  a  limiting  plane  that  cuts  all  the  elements. 

Axis,  a  straight  line  through  the  apex  and  the  center  of  the 
base. 

A  Frustum  is  a  portion  of  the  surface  between  two  parallel 
planes  on  the  same  side  of  the  apex,  which  cut  all  the  elements. 

157.  Discussion. — The  surface  of  the  cone  that  lies  between  the 
base  and  apex  is  usually  all  that  is  considered.     Theoretically, 
however,  the  surface  is  unlimited,   and  extends  beyond  both 
limits    to    infinity.     This    theoretical    extension    creates    two 
divisions  of  the  surfaces,  one  on  each  side  of  the  apex,  called 
respectively  the  Upper  Nappe  and  Lower  Nappe.     In  practical 
work    this    condition    almost    never    exists,    and    is    usually 
ignored. 

Varieties  of  Cones. — The  different  varieties  of  cones  are  dis- 
tinguished by  their  bases  and  axes.  The  bases  may  be  circular, 
elliptical,  hyperbolic,  etc.,  or  combinations  of  curves,  and  the 
axes  are  either  perpendicular  (right)  or  oblique  to  the  bases. 
Thus  we  designate  a  cone  as  a  right  circular  cone,  an  oblique 
elliptical  cone,  etc.  The  base  and  apex,  or  the  base  and  axis,  are 
sufficient  to  determine  the  cone. 

158.  Limits. — A  cone,  whose  apex  is  at  infinity,  becomes  a 
cylinder,  and  one,  whose  apex  is  in  the  plane  of  the  base,  be- 
comes a  plane. 

159.  A  Tangent  Plane  to  a  cone  at  any  point  on  the  surface 
is  tangent    all    along   the    element  through  that  point.      This 
element  is  called  the  Element  of  Contact.     As  all  elements  pass 
through  the  apex,  every  tangent  plane  must  contain  the  apex. 
This  fact  is  of  great  importance  in  passing  tangent  planes.     It 
means  that  any  line  through  the  apex,  except  lines  inside  the 
cone,  will  lie  in  one  or  two  tangent  planes.     From  this  fact,  and 
from  Art.  155,  we  derive  the  special  rule  for  passing  tangent 
planes  to  cones. 

Rule. — One  tangent  line  and  the  apex  are  sufficient  to  determine 
a  tangent  plane  to  any  cone. 


CURVED  LINES  AND  SURFACES  .  123 

Axiom. — If  the  base  of  a  cone  is  in  H,  the  //-trace  of  any  tan- 
gent plane  will  be  tangent  to  the  base. 

16Q.  Development  of  Cones. — When  the  surface  of  a  cone  is 
rolled  out,  we  find  that  there  is  one  point  common  to  all  ele- 
ments, the  apex,  and  this  point  will  remain  a  point  in  the  develop- 
ment. From  this  point  all  the  elements  radiate  in  fan  shape, 
making  up  an  integral  of  all  the  minute  triangles  that  form  the 
surface  of  the  cone.  In  the  case  of  a  right  circular  cone,  the  ele- 
ments are  all  of  equal  length,  and  the  base  rolls  out  into  an  arc 
of  a -circle,  whose  radius  is  the  length  of  the  elements  (sometimes 
called  "slant  height").  The  length  of  the  arc  thus  derived  is 
equal  to  the  circumference  of  the  base.  All  other  cones  have  ele- 
ments of  varying  length,  and  their  bases  will  not  roll  out  in  cir- 
cular paths.  The  development  of  all  cones,  except  the  right 
circular,  is  accomplished  by  "triangulation" ;  that  is,  dividing 
the  surface  into  small  triangles,  and  laying  them  out  flat  in 
order  and  (approximately)  true  size.  An  exception  to  this  may 
be  found  in  an  oblique  elliptical  cone  that  may  have  a  circular 
right  section.  If  such  be  found,  the  circular  method  is  better 
than  triangulation. 

PROBLEMS  RELATING  TO  CONES 

161.  Problem  60. — To  Locate  a  Point  on  the  Surface  of  a  Cone. 
Construction. — See  Fig.  82. 

1.  Assume  either  projection  of  the  point  0,  say  the  //"-projec- 
tion. 

2.  Through  o  and  b(the  apex)  draw  the  //-projection  of  the 
element. 

3.  Project  c  to  c'  on  GL. 

4.  Draw  bV. 

5.  Project  o  to  o'  on  bV. 

162.  Problem  61.— To  find  the  Line  of  Intersection  of  a  Cone 
and  Plane. 

Analysis. — Find  the  piercing  points  of  a  reasonable  number  of 
elements  with  the  given  cone.  The  line  through  these  points 
will  be  the  line  of  intersection. 

Construction. — Shown  in  Fig.  83. 

Note. — In  case  the  piercing  points  are  difficult  to  locate  by  the  ordinary 
process,  the  best  and  easiest  way  to  accomplish  it  is  by  the  method  of 
auxiliary  plane  projection  described  in  Art.  105. 


124 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


In  case  the  plane  is  parallel  to  GL  the  operation  is  easiest  by 
the  use  of  the  profile. 

163.  Problem  62. — To  find  the  Piercing  points  of  a  Line  and  a 
Cone. 

Analysis. — 1.  Pass  a  plane  through  the  given  line  and  the 
apex. 

2.  Obtain  the  elements  cut  by  this  plane. 

3.  Where  the  elements  intersect  the  given  line  will  be  the 
required  piercing  points. 


FIG.  82. — A  point  on  the  surface  of  a  cone,  and  the  tangent  plane  at  that  point. 

Construction. — Let  the  cone  PO  (Fig.  84)  and  the  line  AB  be 
given. 

1.  Pass  a  plane  through  AB  and  P,  find  its  #-trace,  cd.     (V- 
trace  not  necessary.) 

2.  Find  the  points,  e  and  f,  where  cd  intersects  the  base. 

3.  Draw  the  elements,  pe  and  pf. 

4.  Find  the  intersections,  m  and  n,  of  the  elements  and  ab. 

5.  Project  m  and  n  to  m'  and  n'  on  a'b'.     M  and  N  will  be  the 
required  piercing  points. 

Which  operation  determines  whether  AB  will  pierce  the  cone 
or  be  altogether  outside  of  it? 

Note. — There  are  occasions  when  a  plane  cutting  a  number  of  elements 
can  be  used  to  better  advantage  than  a  plane  through  the  apex.  Such  a 
plane  will  cut  a  curve  from  the  cone  surface,  and  the  piercing  points  will  be 
at  the  intersections  of  the  given  line  with  the  curve. 


CURVED  LINES  AND  SURFACES 


125 


FIG.  83. — A  cone  intersected  by  a  plane. 
P" 


FIG.  84. — Piercing  points  of  a  line  and  a  cone. 


126  PRACTICAL  DESCRIPTIVE  GEOMETRY 

164.  If  the  base  of  the  cone  is  not  in  H,  V  or  P}  the  problem 
is  more  complicated,  but  the  method  is  unchanged.     The  student 
must  accustom  himself  to  problems  that  involve  new  applica- 
1<ions  and  changed  conditions  of  the  processes  illustrated. 

165.  Problem  63. — To  pass  a  plane  Tangent  to  a  Cone  at  a 
point  on  the  surface. 

Analysis. — Read  the  rule  given  in  Art.  159. 
Construction.— Refer  to  Fig.  82,  Art.  162.     Let  O  be  the  given 
point. 

1.  Draw  the   element   through   O.     This   is   the  Element  of 
Contact,  and  one  line  of  the  plane. 

2.  Where  the  element  of  contact  intersects  the  base,  draw  the 
tangent  to  the  base. 

This  is  the  JET-trace  of  the  required  plane,  and  the  F-trace  may 
be  obtained  in  several  ways  known  to  the  student. 

166.  Problem  64. — To  pass  a  plane  Tangent  to  a  Cone  through 
a  Point  outside  the  surface. 

Analysis. — 1.  Draw  a  line  through  the  apex  and  the  given 
point.  (See  Art.  159.) 

2.  Find  the  piercing  point  of  this  line  with  the  plane  of  the 
base. 

3.  From  this  point  draw  one  of  the  two  possible  tangents  to 
the  base. 

4.  Pass  the  required  plane  through  the  tangent  line  and  the 
apex. 

Construction. — Let  the  student  make  it. 

Note  1. — If  the  line  through  the  given  point  and  the  apex  is  parallel  to 
the  base,  the  tangent  line  in  the  plane  of  the  base  will  be  parallel  to  said 
line,  and  may  be  drawn  on  either  side  of  the  base. 

Note  2. — If  the  line  through  the  apex  is  too  nearly  parallel  to  the  plane  of 
the  base  to  obtain  the  piercing  point  within  the  limits  of  the  drawing,  a 
plane  may  be  passed  parallel  to  the  base,  nearer  the  apex,  and  a  tangent 
may  be  drawn  to  the  new  base. 

167.  Problem  65. — To  pass  a  plane  Tangent  to  a  Cone  parallel 
to  a  given  line. 

Analysis. — 1.  Draw  a  line  through  the  apex  parallel  to  the 
given  line. 

2.   Having  this  line,  finish  as  in  Problem  64. 

Construction. — Let  the  student  make  it. 

Impossible  Conditions. — When  the  line  through  the  apex 
parallel  to  the  given  line  is  inside  the  cone,  the  problem  is  impos- 


CURVED  LINES  AND  SURFACES 


127 


sible.  In  such  a  case  the  line  will  pierce  the  plane  of  the  base 
inside  the  base.  Otherwise  the  problem  is  possible. 

168.  Problem  66. — To  pass  a  plane  Tangent  to  a  Cone,  making 
a  given  angle  with  H. 

Limitations. — Only  certain  conditions  will  allow  a  solution  of 
this  problem. 

1.  If  the  cone  be  a  right  circular  cone,  with  its  base  on  H, 
there  is  only  one  angle  possible,  and  that  is  the  anlge  that  its 
elements  make  with  H. 


FIG    85. — Plane  tangent  to  a  cone,  and  inclined  60°  to  the  horizontal. 

2.  If  the  base  be  in  some  plane  not  parallel  to  H}  there  are 
many  angles  at  which  tangent  planes  may  be  passed,  though  not 
all. 

3.  If  the  base  be  in  H,  and  the  axis  oblique,  different  angles, 
but  not  all,  may  be  assigned  to  the  problem. 

4.  If  the  base  be  in  H,  and  a  different  curve  from  the  circle, 
even  though  the  axis  be  perpendicular,  there  will  be  some  variety 
to  the  angles  that  may  be  required. 

Each  problem  will  fix  its  own  limitations. 

Construction. — Let  it  be  required  to  pass  a  plane  tangent  to 
the  cone  AB  (Fig.  85)  making  60°  with  H. 

1.  With  B  as  its  apex,  draw  the  projections  of  a  60°  cone, 
axis  BC. 


128  PRACTICAL  DESCRIPTIVE  GEOMETRY 

2.  Draw  the  H-trgice  of  T,  tangent  to  the  bases  of  the  two 
cones.     This  may  be  drawn  on  either  side  of  the  cones. 

3.  Locate  the  F-trace  of  T  by  any  lines  through  B,  lying  in  T. 

Note. — Under  certain  conditions  four  planes  can  be  passed  making  the 
same  angle  with  H,  all  tangent  to  the  cone. 

169.  Problem  67. — To  develop  the  surface  of  a  right  circular 
cone. 

Analysis.— Read  Art.  160. 

Construction  — Let  it  be  required  to  develop  the  truncated 
cone  in  Fig.  83;  that  is,  the  surface  included  between  the  base 
and  the  plane  T. 

1.  Divide  the  base  into  any  convenient  number  of  equal  partsr 
and  draw  the  elements. 

2.  From  b',  the  apex,  on  the  F-projection,  describe  an  arc  of 
a  radius  equal  to  the  true  length  of  the  elements. 

3.  Make  this  arc  equal  in  length  to  the  circumference  of  the 
base  of  the  cone,  and  divide  it  into  the  same  number  of  equal 
parts.     (See  Note  below.) 

4.  Measure  the  true  length  of  the  truncated  elements,  and  lay 
them  off  on  the  corresponding  radii  of  the  Pattern  (development) . 

5.  Draw  a  smooth  curve  through  these  points. 

Note. — The  accurate  way  to  lay  off  this  arc  equal  in  length  to  the  circum- 

R     360°  r 

ferenceis  by  this  equation :  — ==  —E$-,  or  #0  =  -n  360°,  where  R  equals  the  slant 

height  of  the  cone,  r  the  radius  of  the  base,  and  6  is  the  number  of  degrees 
in  the  angle  of  the  development.  It  can  be  done  more  roughly  by  step- 
ping off  small  arcs  with  the  dividers. 

170.  Problem  68. — To  develop  the  surface  of  an  oblique  cone. 

This  is  accomplished  by  triangulation,  and  it  is  as  nearly  exact 
as  the  chord  of  a  small  arc  is  to  the  length  of  the  arc. 
Construction.— The  cone  PO  (Fig.  86). 

1.  Divide  the  base  into  a  number  of  equal  arcs. 

2.  Draw  the  elements  of  these  divisions,  dividing  the  cone 
into  small  (approximate)  triangles. 

3.  Measure  the  true  length  of  all  the  elements. 

Note. — By  a  little  care  in  the  division  this  can  be  done  in  pairs. 

4.  Rectify  one  of  the  equal  arcs. 

5.  Beginning    (preferably)    with    the    longest    element,    01, 
construct  triangles  equal  to  the  (approximate)  true  size  of  those 
in  the  cone,  as  shown  in  the  development  at  the  right. 


CURVED  LINES  AND  SURFACES 


129 


Second  Method. — A  method  often  used  in  the  development  of 
oblique  cones  in  the  "Sphere  Method."  The  process  is  as 
follows: 

1.  Draw  the  projections  of  a  sphere,  of  any  convenient  size, 
with  its  center  at  the  apex. 

2.  Obtain  the  piercing  points  of  the  various  elements  on  the 
surface  of  the  sphere. 

3.  The  segments  of  the  elements  thus  obtained  are  all  of  equal 
length,  and  will  therefore  be  developed  on  the  arc  of  a  circle, 
like  the  right  circular  cone. 


Development. 
FIG.  86. — The  shortest  line  between  two  points  on  the  surface  of  a  cone. 


4.  The  remainder  of  each  element  can  then  be  measured,  and 
laid  off  on  the  development  in  its  proper  place. 

Let  the  student  make  the  construction. 

171.  The  Problem  of  the  Shortest  Path. 

The  shortest  line  that  can  be  drawn  on  any  surface  from  one 
point  to  another  is  a  straight  line  on  the  development.  Let 
A  and  B  (Fig.  86)  be  two  points  on  the  surface  of  the  cone. 

1.  Locate  them  on  the  development,  see  a1  and  bt. 

2.  Draw  a  straight  line  from  at  to  bx,  and  locate  its  intersec- 
tions, c1?  d1?  etc.,  with  the  intervening  elements. 


130  PRACTICAL  DESCRIPTIVE  GEOMETRY 

3.  Return  all  these  points  to  their  original  positions  in  the 
projections  of  the  cone,  and  draw  the  curve  through  them. 

Note.  —The  shortest  path  on  a  cylinder  is  always  a  helix,  a  circle,  or  a 
straight  line. 

172.  EXERCISES 

531.  Find  the  points  in  which  A(l+2-i)B(4  +  $-3)    pierces   the    cone, 
whose  axis  is  C(1J,  0-2J)  D(3|  +  3-lf),  and  base  a  2  in.  circle  in  H. 

532.  Draw  the  projections  of  a  right  circular  cone,  tangent  to  H,  its  axis 
A(2,  0-1)   B(4  +  l-2i).      Find  where  it  is  pierced  by  C(2  +  2-2}) 
D(4,  0-li). 

533.  Draw   the   projections    of   a   right    circular   cone,    axis   A(2  +  l  —  1^) 
B(4  +  l  —  1£),  base  a  If-in.  circle  at  B.     Find  where  it  is  pierced  by 
C(2  +  i-i)  D(4  +  2-2). 

534.  Draw  the  shortest  line  that  can  be  drawn  on  the  cone  in  Ex.  533  from 
E(3  +  f,  x)  on  the  rear  surface  to  F(3|  +  l|,  y)  on  the  front  surface. 

535.  Draw  the   shortest   line    from  M(2£  +  £,  x)  on   the  rear    surface    to 
N  (3J  +  1,  y)  on  the  front  surface  of  the  cone  in  Ex.  532. 

536.  Find  the  intersection  of  the  plane  T(l+2)   5(3-3)  with  the  right 
circular  cone  of  2-in.  base,  whose  axis   is  A(3,  0  —  1)   B  (3  +  2^  —  1). 
Find  the  true  size  of  the  section. 

537.  Develop  the  cone  in  Ex.  536  between  H  and  T. 

538.  Given  the  cone  in  Ex.  536.     Intersect  it  by  four  planes,  all  perpen- 
dicular to  V,  that  will  cut  it  respectively  in  a  circle,  ellipse,  hyperbola 
and  parabola.     Show  each  curve  in  its  true  size. 

539.  Let  A(l$  +  2i  —  2)  B(4  +  l  —  1)  be  the  axis  of  a  right  circular  cone,  of 
2-in.  base,  apex  at  A.     Draw  its  projections.     Find  the  intersection 
with  the  plane  T(4  +  3)  1(4-3). 

Note.  —  This  can  be  much  easiest  worked  by  the  auxiliary  plane  method 
described  in  Art.  105. 

540.  Develop  the  surface  of  the  cone  in  Ex.  539  between  the  apex  and  T. 

541.  M(2  +  £  —  2)  N  (3  +  2-l|)  is  the  axis  of  a  right  circular  cone  of  l$-in. 
diameter  at  M.     Find  its  intersection  with  H. 

542.  Develop  the  surface  of  the  cone  in  Ex.  541  between  the  apex  and  H. 

543.  A(3  +  l  —  i)   B(3  +  l—  2|)  is  the  axis  of  a  right  circular  cone,  base  a 
2-in.  circle  at  A.     Draw  the  traces  of  a  plane  tangent  at  O(2$  +  f,  x) 
on  the  surface. 

544.  Draw  the  traces  of  a  plane,  tangent  to  the  cone  in  Ex.  543,  containing 


545.  Draw  the  traces  of  a  plane,  tangent  to  the  cone  in  Ex.  543,  parallel  to 


546.  Draw  the  traces  of  a  plane,  making  60°  with  H.,  tangent  ,to  the  cone  in 
Ex.  543, 

547.  Draw  the  traces  of  the  plane,  tangent  to  the  cone  in  Ex.  532,  containing 
the  point  C(3  +  f  ,  x)  on  the  rear  surface. 

548.  Draw  the  traces  of  a  plane,  tangent  to  the  cone  in  Ex.  533,  containing 
the  point  N(2  +  i-l). 


CURVED  LINES  AND  SURFACES  131 

549.  Draw  the  traces  of  a  plane,  tangent  to  the  cone  in  Ex.  533,  parallel  to 
O(l  +  2-3)  P(2  +  l-li). 

550.  Draw  the  traces  of  a  plane,  tangent  to  the  cone  in  Ex.  533,  making 
45°  with  H. 

551.  Draw  a  right  circular  cone,  axis  A(3  +  f  —  f)  B(3  +  2  —  1£),  base  a  l£-m. 
circle  at  A.     Draw  the  traces  of  a  plane  tangent  at  O(2f  +  1,  x)  on  the 
surface. 

552.  Draw  the  traces  of  a  plane,  tangent  to  the  cone  in  Ex.  551,  containing 
the  point  P  (2  +  1-  1$). 

553.  Draw  the  traces  of  a  plane,  tangent  to  the  cone  in  Ex.  551,  parallel  to 
A(l  +  2-3)  B(2  +  l-l£). 

554.  Draw  the  traces  of  a  plane,  tangent  to  the  cone  in  Ex.  539,  containing 

KCB  +  J-1). 

555.  Draw  the  traces  of  a  plane,  tangent  to  the  cone  in  Ex.  539,  containing 
E(3  +  l,  x)  on  the  rear  surface. 

556.  Draw  the  traces  of  the  plane,  tangent  to  the  cone  in  Ex.  532,  containing 
the  point  C(3  +  f,  x)  on  the  rear  surface. 

557.  Draw  the  traces  of  a  plane,  tangent  to  the  cone  in  Ex.  532,  containing 
the  point  F(4  +  $-l). 

558.  Given  an  oblique  cone,  axis  A(2$,  0-2)   B(4  +  2-l),  base  a  2-in. 
circle  in  H.     Draw  the  traces  of  the  plane  tangent  at  M(2,  x  —  2$)  on 
the  surface. 

559.  Draw  the  traces  of  a  plane,  tangent  to  the  cone  in  Ex.  558,  containing 


560.  Draw  the  traces  of  a  plane,  tangent  to  the  cone  in  Ex.  558,  parallel  to 
X(2  +  3,  0)  Y(l  +  l-3). 

561.  Draw  the  traces  of  a  plane  inclined  75°  to  H,  tangent  to  the  cone  in 
Ex.  558.     How  many  such  planes  are  possible? 

562.  Draw  a  right  circular  cone  tangent  to  H,  axis  M(2  +  1  -  1$)  N(4,  0-  1$). 
Draw  the  traces  of  a  tangent  plane  through  O(3  +  f,  x)  on  the  rear 
surface. 

563.  Draw  the  traces  of  a  plane,  tangent  to  the  cone  in  Ex.  562,  containing 


564.  Given  a  right  circular  cone,  axis  A(2,  0  —  1$)  B(2  +  2£-l£),  base  2-in. 
diameter  at  A.     Place  a  1^-in.  sphere  so  as  to  be  tangent  to  H,  V  and 
the  cone.     Draw  the  traces  of  a  plane  tangent  to  the  sphere  and  cone, 
but  not  between  them. 

565.  Draw  the  traces  of  a  plane  tangent  to  two  right  circular  cones,  both  of 
whose  apices  are  in  the  point  M(3  +  l  —  1$),  one  having  a  2-in.  base  in 
V,  and  the  other  a  1-in.  base  in  H. 

173.  PRACTICAL  EXERCISES 

566.  Draw  the  patterns  (development)  for  the  funnel   (Fig.  87)   to  a  con- 
venient scale. 

567.  Draw  the  patterns  for  the  ventilator  hood  and  deflector  (Fig.  87)  to 
scale.     Use  two  spaces. 

568.  Draw  the  patterns  for  the  conical  offset  pipe  (Fig.  87)  to  any  convenient 
scale. 


132 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Conical  Offset  Pipe  --  14-" - 

Conical  Elbow. 


k-—  12"- — 


Furnace  Hood. 


Reducer. 


FIG.  87. — Practical  examples  of  conical  surfaces. 


CURVED  LINES  AND  SURFACES  133 

569.  Lay  out  the  patterns  for  the  sink  drainer  (Fig.  87)  to  any  convenient 
scale. 

570.  The  conical  elbow,  shown  in  Fig.  87,  is  composed  of  two  cones,  the  right 
sections  of  the  larger  being  circles.     The  smaller  has  an  elliptical 
lower  base,  and  the  upper  base  to  be  determined.     Draw  both  pro- 
jections, find  out  the  true  shape  of  the  spout,  and  lay  out  the  patterns 
to  scale.     Use  two  spaces. 

571.  Develop  the  conical  furnace  hood    (Fig.  87)  along  a  "  spiral  seam." 
Make  the  plan  of  the  conical  helix  a  Spiral  of  Archimedes,  making  one 
complete   revolution.     Omit   lap   and   rivet   holes.     Any   convenient 
scale. 

572.  Lay  out  the  patterns  for  the  chimney  reducer,  shown  in  Fig.  87,  to 
any  convenient  scale. 

THE  CYLINDER 

174.  Definition. — A  cylinder  is  a  single-curved  surface  gene- 
rated by  a  line  moving  along  any  curved-line  directrix,  always 
remaining  parallel  to  a  given  line.     It  is  a  cone  with  its  apex  at 
infinity. 

The  Surface  of  a  cylinder  is  usually  taken  to  be  that  portion 
which  is  included  between  two  plane  bases.  The  bases  may  be 
parallel  or  not.  Theoretically  the  cylinder  is  infinite  in  extent. 

Axis. — The  axis  is  a  line  through  the  center  of  a  right  section, 
parallel  to  all  the  elements.  It  is  a  useful  line  in  defining  the 
cylinder  in  space. 

Varieties  of  Cylinders. — The  characteristic  distinction  between 
cylinders  is  the  curve  of  their  right  section.  This  is  at  some 
variance  to  the  cone  distinctions,  which  are  founded  on  base 
and  axial  peculiarities.  The  reason  for  this  is  that  it  is  absolutely 
necessary  to  know  the  right  section  of  a  cylinder  in  most  of  the 
problems  arising;  as,  for  instance,  in  its  development,  the  right 
section  always  rolls  out  in  a  straight  line.  Also,  in  most  tangent 
problems,  the  right  section  has  to  be  considered.  So,  when  a 
cylinder  is  spoken  of  as  circular,  elliptical,  etc.,  a  right  section 
is  understood. 

Limits. — When  the  directrix  becomes  a  straight  line,  the 
cylinder  becomes  a  plane,  and  when  the  directrix  becomes  a 
point  the  cylinder  becomes  a  line. 

175.  Tangent  Planes  to  Cylinders. — A  plane  which  contains 
two   consecutive   elements   is   tangent.     The  Tangent  Plane   is 
parallel  to  the  axis,  and  to  every  element.     The  two  consecutive 
elements  contained  in  the  tangent  plane  form  the  Element  of 
Contact.     This  element  of  contact  is  one  of  the  two  tangent  lines 


134  PRACTICAL  DESCRIPTIVE  GEOMETRY 

necessary  to  determine  the  tangent  plane.  From  this  fact  we 
derive  the  special  rule  for  passing  planes  tangent  to  cylinders. 

Rule. — A  tangent  line  and  a  line  parallel  to  the  axis,  inter- 
secting the  tangent,  determine  a  tangent  plane  to  a  cylinder. 

Axiom. — When  the  base  of  a  cylinder,  is  in  H ,  the  .H-trace  of 
any  plane,  tangent  to  the  cylinder,  will  be  tangent  to  the  base 
curve  of  the  cylinder. 

Axiom. — A  plane  cannot  in  general  be  tangent  to  two  cylinders 
unless  the  axes  are  parallel.  The  exceptions,  which  are  not 
frequent,  will  only  be  found  when  two  outside  elements  of  the 
two  cylinders  are  in  the  same  plane.  That  plane  will  be  the 
tangent  plane.  If  two  cylinders  have  parallel  axes,  four  planes 
tangent  to  both  can  usually  be  passed.  Two  cylinders  of  equal 
diameter,  with  axes  intersecting,  can  have  two  common  tangent 
planes  outside. 

176.  Development. — As    was    before    mentioned,    the    right 
section  of  a  cylinder  rolls  out  into  a  straight  line.     This  is  true, 
because  the  right  section  is  perpendicular  to  all  the  elements,  and 
when  they  are  rolled  out  they  remain  parallel,  and  the  plane 
perpendicular  to  them  cuts  their  plane  in  a  straight  line   per- 
pendicular to  them.     Therefore,  the  first  thing  to  do  is  to  rectify 
a  right  section.     Then  lay  off  perpendiculars,  properly  spaced, 
with  the  true  lengths  of  the  various  elements  laid  off  in  order. 

Axiom. — The  cylinder  included  between  two  right  sections 
becomes  a  rectangle  in  development. 

PROBLEMS  RELATING  TO  CYLINDERS 

177.  Problem  69. — To  locate  a  point  on  the  surface. 
Analysis. — 1.  Assume  one  projection  of  the  point. 

2.  Draw  a  line  through  that  projection  parallel  to  the  axis. 
(This  will  be  the  element  on  which  it  lies.) 

3.  Find  the  intersection  of  this  element  with  the  base. 

4.  Project  this  intersection  to  the  other  projection  of  the  base 
and  draw  the  other  projection  of  the  element. 

5.  Project  the  point  to  the  second  projection  of  the  element. 
Let  the  student  make  the  construction. 

178.  Problem  70. — To  find  the  intersection  of  a  plane  and  a 
cylinder. 

Analysis. — 1.  Draw  a  sufficient  number  of  elements. 

2.  Obtain  the  piercing  points  of  these  elements  with  the  plane. 

3.  Draw  a  smooth  curve  through  these  piercing  points  in  order. 


CURVED  LINES  AND  SURFACES  135 

179.  Problem  71. — To  find  the  points  in  which  a  straight  line 
pierces  a  cylinder. 

There  are  two  methods:  1.  to  use  when  the  base  of  the  cylinder 
lies  in  one,of  the  planes  of  projection;  2.  to  use  when  the  base  is 
in  space. 

Analysis.    First  Method.—  When  the  base  is  in  H,  V  or  P. 

1.  Pass  a  plane  through  the  line  parallel  to  the  axis. 

2.  At  the  points  in  which  the  trace  of  this  plane  crosses  the 
base,  draw  elements  of  the  cylinder. 

3.  The  points  in  which  these  elements  are  intersected  by  the 
given  line  will  be  the  required  points. 

Second  Method. — When  the  base  is  in  space. 

1.  Draw  a  sufficient  number  of  elements. 

2.  Pass  a  plane  through  the  line,  preferably  a  projecting  plane. 

3.  Find  the  intersections  of  this  plane  with  enough  of  the 
elements  to  yield  a  curve  that  will  be  intersected  twice  by  the 
given  line. 

4.  The  required  points  will  then  be  where  the  given  line  crosses 
the  curve. 

180.  Problem  72. — To  pass  a  Plane  Tangent  to  a  Cylinder 
through  a  Given  Point  on  the  Surface. 

Analysis. — 1.  Through  the  given  point  draw  a  line  parallel  to 
the  axis. 

(This  will  be  one  line  of  the  required  plane,  the  element  of 
contact.) 

2.  At  the  point    where  the  element  of  contact  crosses  any 
curve  of  the  surface   (preferably  one  of  the  bases),  draw  the 
tangent  to  that  curve. 

3.  Pass  the  plane  through  the  element  and  tangent. 
Construction  — Let  O   (Fig.  88)    be  the   given   point    on   the 

surface  of  the  elliptical  cylinder  whose  axis  is  AB. 

1.  Draw  the  element  OP. 

2.  Draw  the  tangent  to  the  base  at  P. 

3.  This  tangent  will  be  the  H -trace  of  the  tangent  plane. 

4.  Draw  the  F-trace  by  any  of  the  known  means. 

Note. — This  construction,  as  illustrated  in  Fig.  88,  is  only  possible  when 
the  base  of  the  cylinder  is  in  H,  V  or  P.  If  the  base  is  in  an  oblique  plane, 
the  tangent  will  not  be  one  of  the  traces  of  the  plane.  The  analysis  will 
apply  to  any  situation  of  the  cylinder. 

181.  Problem  73. — To  pass  a  Plane  Tangent  to   a  Cylinder 
through  a  Point  outside  the  surface. 


136 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Analysis. — 1.  Draw  a  line  through  the  point  parallel  to  the  axis. 

2.  Find  the  point  in  which  this  line  pierces  the  plane  of  the 
base. 

3.  From  this  piercing  point  draw  a  tangent  to  the  base. 

4.  The  parallel  line  and  the  tangent  line  will  determine  the 
required  tangent  plane. 

Note. — Let  the  student  assume  an  outside  point  and  make  his  own  con- 
struction for  a  cylinder  like  the  one  illustrated  in  Fig.  88.  A  more  difficult 
construction  is  shown  in  Fig.  88a. 


FIG.  88. 

Construction. — Let  the  cylinder  be  a  circular  cylinder  of  f-in. 
diameter,  whose  axis  is  A  B  (Fig.  88a).  Let  it  be  required  to 
pass  a  tangent  plane  through  M,  a  point  outside  the  surface. 

1.  Draw  the  line  MN,  parallel  to  AB  through  M. 

2.  Pass  a  plane  Q  through  A,  perpendicular  to  AB.     (This 
will  be  the  plane  of  the  base,  cutting  a  circle  from  the  surface.) 

3.  Find  the  piercing  point,  N,  of  MN  with  Q. 

4.  Revolve  A  and  N  into  H  about  Qq. 

5.  Draw  a  J-in.  circle  about  &±  as  a  center. 

6.  From  nx  draw  a  tangent  to  the  circle. 

7.  Counter-revolve  the  tangent  n'o  into  the  plane  Q. 

8.  Pass  the  plane  S  through  NO  and  MN.     It  will  be  the 
required  tangent  plane. 


CURVED  LINES  AND  SURFACES 


137 


Note. — Only  the  axis  of  this  cylinder  is  drawn  as  the  problem  can  be 
worked  without  drawing  the  projections  of  the  entire  cylinder.  To  show  it 
would  unnecessarily  complicate  the  figure.  The  student  should  familiarize 
himself  with  this  sort  of  problem,  and  will  find  examples  of  it  in  Exercises 
603-614,  620-627. 

182.  Problem  74.— To  pass  a  Plane  Tangent  to  a  Cylinder  and 
parallel  to  a  given  line. 

Analysis. — 1.  Pass  a  plane  parallel  to  the  axis  and  the  given 
line.  Why? 


FIG.  88a. 

2.  Find  the  intersection  of  this  plane  and  the  plane  of  the  base. 

3.  Draw  a  tangent  to  the  base  parallel  to  the  intersection 
obtained  in  operation  (2) . 

4.  Pass  the  required  tangent  plane  through  this  tangent  line, 
parallel  to  the  auxiliary  plane,  passed  in  operation  (1). 

Construction. — Let  AB,  Fig.  89,  be  the  axis  of  the  cylinder, 
and  let  it  be  required  to  pass  the  tangent  plane  parallel  to  XY. 

1.  Pass  the  auxiliary  plane  Q  through  AB  parallel  to  XY. 

2.  Draw  the  required  plane  T,  tangent  to  the  cylinder  and 
parallel  to  Q.     The  //"-trace  of  T  will  be  tangent  to  the  base  in 
H ,  and  parallel  to  Qq,  and  the  F-trace  of  T  will  be  parallel  to  Qq'. 

183. — If  the  cylinder  be  circular  and  its  axis  oblique,  the  best 
method  for  the  foregoing  problem  is  the  "Normal"  method. 
Its  analysis  is  as  follows: 

1.  Pass  a  plane  through  the  axis  parallel  to  the  given  line. 


138 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


FIG.  89. 


CURVED  LINES  AND  SURFACES 


139 


2.  Pass  a  plane  parallel  to  this  plane  at  a  distance  equal  to  the 
radius  of  the  cylinder  (Problem  34). 

The  problem  may  be  done  in  this  way  without  drawing  any 
elements  or  bases  of  the  cylinder.  Let  the  student  prove  that 
the  " normal"  method  is  correct. 

184.  Problem  75.  To  pass  a  Plane,  inclined  at  any  given  angle 
with  H  or  V,  tangent  to  a  Cylinder. 

Limitations. — The  limits  of  this  angle  must  be  between  90° 
and  the  angle  between  the  elements  and  the  stipulated  plane. 

Analysis. — 1.  Pass  an  auxiliary  plane  through  the  axis  making 
the  given  angle  with  H  or  V,  as  required;  see  Problem  40. 


Development. 


FIG.  91. 


2.  If  the  base  of  the  cylinder  is  in  H,  draw  the  #-trace  of  the 
required  plane,  tangent  to  the  base  and  parallel  to  the  H-trace 
of  the  auxiliary  plane. 

3.  Draw  the  F-trace  of  the  required  plane  parallel  to  the  V- 
trace  of  the  auxiliary  plane. 

185.  Normal  Method. — When  this  problem  is  required  of  a 
circular  cylinder,  oblique  to  H  and  Vt  the  normal  method,  like 
that  in  Art.  183,  is  the  easiest  and  best. 

Fig.  90  shows  the  construction  of  such  a  problem,  in  which 
AB  is  the  axis  and  AC  the  radius  of  the  given  cylinder.  The 
elements  of  the  auxiliary  cone  make  the  given  angle  with  H. 
T  is  one  of  the  four  possible  required  planes. 

186.  When  the  axis  is  parallel  to  H,  and  the  angle  between 


140  PRACTICAL  DESCRIPTIVE  GEOMETRY 

the  tangent  plane  and  H  is  given,  the  problem  is  very  easily 
solved  by  projecting  on  a  plane  perpendicular  to  the  axis. 

Note. — Four  planes  can  be  passed  tangent  to  any  cylinder,  making  the 
given  angle,  if  the  problem  is  possible. 

187.  Problem  76.— To  Develop  the  Surface  of  a  Cylinder. 
Construction. — To  develop  the  elliptical  cylinder,  whose  axis 

is  AB  (Fig.  91). 

1.  Draw  a  sufficient  number  of  elements. 

2.  Pass  a  plane  perpendicular  to  the  axis,  as,  for  example, 
along  the  line  CD. 

3.  Revolve  this  right  section  to  show  its  true  size. 

4.  Lay  off  the  base  line,  Cfl^  equal  in  length  to  the  length  of 
the  ellipse,  with  divisions  equal  to  the  respective  arcs,  as  shown 
by  the  corresponding  numerals. 

f>.  Through  all  these  divisions  draw  perpendiculars  to  CjD^ 

6.  Lay  off  on  these  perpendiculars  the  true  lengths  of  their 
respective  elements,  measured  above  and  below  the  right  section. 

7.  Draw  a  smooth  curve  through  the  extremities. 

Note. — It  is  not  always  necessary  to  show  both  projections  of  the    right 
section.     In  this  case  the    //-projection  is  unnecessary. 

EXERCISES 

188.  Where  elements  are  needed,  use  at  least  16  in  these  exercises. 

573.  Find  the  line  of  intersection,  and  its  true  size,  of  the  plane  T(l+3) 
5(1—3)    and   the   2-in.    circular  cylinder,    whose   axis  is  A(2,   0  —  1) 
B(2  +  3-l). 

574.  Find  the  line  of  intersection,  and  its  true  size,  of  the  plane  S(  +  2)oo 
( —  3)  with  the  cylinder  in  Exercise  573. 

575.  Find  the  intersection,  and  its  true  size,  of  the  plane  W(l+2)  5(1—3) 
with  the  elliptical  cylinder,  axis  C(l,  0-1)  D(3  +  3-2),  base  a  2-in. 
circle  in  H. 

576.  Find  the  intersection,  and  its  true  size,  of  the  plane  X(l+2)  3(3  —  3) 
and  the  cylinder  in  Exercise  575. 

577.  Find  the  true  size  of  a  right  section  of  the  cylinder  in  Exercise  575. 

578.  Find  the  intersection,  and  its  true  size,  of  the  plane  T(3  +  3)  5(3  —  3) 
with  the  1^  in.  circular  cylinder,  whose  axis  is  M(l  +  l  —  1)  N(4  +  3  — 1). 

579.  Find  the  intersection,  and  its  true  size,  of  the  plane  P(3  +  3)  3(3  —  3) 
and  the  cylinder  in  Exercise  575. 

580.  Find  the  intersection,  and  its  true  size,  of  the  plane  S(4  +  3)  2(2-3) 
and  the  circular  cylinder,  tangent  to  H,  axis  K(l  +  f  —  2)  L(4  +  f  —  $). 

581.  Find  the  intersection,  and  its  true  size,  of  the  plane  W(4  +  3)  1(4  —  3) 
and  the  cylinder  in  Exercise  580. 


CURVED  LINES  AND  SURFACES  141 

582.  Find  the  intersection,  and  its  true  size,  of  the  plane  P(l+3)  4(1-2) 
and  the  H-in.  circular  cylinder,  whose  axis  is  A(l  +  1  —  1)  B(5  +  1  —  1). 

583.  Find  the  true  size  of  the  right  section  of  an  elliptical  cylinder,  whose 
axis  is  C(3,  0-2)  D(3  +  3,  0),  and  whose  base  is  a  IJ-m-  circle  in  H. 
Find  its  intersection  with  V. 

584.  Find  the  intersection,  and  its  true  size,  of  the  plane  R(  +  3)oo(  —  2)  and 
the   li-m.   circular  cylinder,   whose  axis  is  E(3  +  l-l)   F(3  +  2  —  3). 
Find  also  the  H  and  V  intersections  of  the  cylinder. 

585.  Find  the  H  and  V  intersections  of  the  1^-in.  circular  cylinder,  whose 
axis  is  G(3,  0-2)  H(3  +  H,  0). 

586.  Find  the  H  and  V  intersections  of  the  1-in.  circular  cylinder,  whose 
axis  is  K(2  +  |-2)  L(4  +  l£-i). 

587.  Find  the  points  in  which  A(l+2  —  3)  B(3  +  l  —  1)  pierces  the  cylinder 
in  Exercise  575. 

588.  Find  the  points  in  which  C(2  +  3-$)  D(3  +  l-2)  pierces  the  cylinder 
in  Exercise  578. 

589.  Find  the  points  in  which  E(2  +  2-2)  F(4  +  i  —  i)  pierces  the  cylinder 
in  Exercise  585. 

590.  Find  the  shortest  line  that  can  be  drawn  from  G(1J  +  1,  x)  on  the 
rear  surface  to  H  (2  ^  +  2,  y)  on  the  front  surface  of  the  cylinder  in 
Exercise  573.     What  is  this  line? 

591.  Find  the  projections  of  the  shortest  line  that  can  be  drawn  on  the 
surface  of  the  cylinder  in  Exercise  580,  from  M(l$,  x—  1J)  on  the  under 
to  N(3,  y  —  H)  on  the  upper  surface.     What  is  this  line? 

592.  Draw  the  shortest  line  from  0(3$,  +  1,  x)  on  the  rear  to  P(2J  +  1J,  y) 
on  the  front  surface  of  the  cylinder  in  Exercise  583. 

593.  Develop  the  surface  of  the  cylinder  in  Exercise  573,  between  H  and  T. 

594.  Develop  the  surface  of  the  cylinder  in  Ex.  574  between  H  and  S. 

595.  Develop  the  surface  of  the  cylinder  in  Ex.  583  between  H  and  V. 

596.  Develop  the  surface  of  the  cylinder  in  Ex.  575  between  H  and  W. 

597.  Develop  the  surface  of  the  cylinder  in  Ex.  584  between  the  base  at  E 
and  the  plane  R. 

598.  Develop  the  surface  of  the  cylinder  in  Ex.  584  between  H  and  R. 

599.  Cylinder  in  Ex.  575.     Draw  the  traces  of  a  tangent  plane  through 
A(2  +  l,  x)  on  the  front  surface. 

600.  Cylinder  in  Ex.  575.     Draw  the  traces  of  a  tangent  plane  through  the 
point  B(2i  +  l-l). 

601.  Cylinder  in  Ex.   575.     Draw  the  traces  of  a  tangent  plane  that  is 
parallel  to  M(5  +  2-2)  N(4,  0-1). 

602.  Cylinder  in  Ex.  575.     Draw  the  traces  of  a  tangent  plane  inclined 
75°  to  H. 

603.  Cylinder  in  Ex.  578.     Draw  the  traces  of  a  tangent  plane  containing 
E(2  +  l,  x)  on  the  front  surface. 

604.  Cylinder  in  Ex.  578.     Draw  the  traces  of  a  tangent  plane  containing 


605.  Cylinder  in  Ex.  578.     Draw  the  traces  of  a  tangent  plane  parallel  to 
K(5  +  2-l)  L(5|  +  l-i). 

606.  Cylinder  in  Ex.  578.     Draw  the  traces  of  a  tangent  plane  inclined  45° 
toff. 


142  PRACTICAL  DESCRIPTIVE  GEOMETRY 

607.  Cylinder  in  Ex.  580.     Draw  the  traces  of  a  tangent  plane  containing 
A(2£  +  f,  x)  on  the  rear  surface. 

608.  Cylinder   in    Ex.    580.      Draw  the    traces    of    both    tangent    planes 
through  B(3  +  2$  —  1£),  and  measure  the  angle  between  them. 

609.  Cylinder  in  Ex.  580.     Draw  the  traces  of  the  four  tangent  planes 
inclined  60°  to  H. 

610.  Cylinder  in  Ex.  580.     Draw  the  traces  of  both  tangent  planes  parallel 
toC(l+2$-2i)  D(H  +  3-3). 

611.  Cylinder  in  Ex.  582.     Draw  the  traces  of  the  tangent  plane  containing 
E(2  +  f,  x)  on  the  lower  rear  surface. 

612.  Cylinder  in  Ex.  582.     Draw  the  traces  of  both  tangent  planes  through 
the  point  F(3  +  2i-$). 

613.  Cylinder  in  Ex.  582.     Draw  the  traces  of  both  tangent  planes  parallel 
to  G(4  +  2-3)  H(4$  +  2f-2). 

614.  Cylinder  in  Ex.   582.     Draw  the  traces  of  the  four  tangent  planes 
inclined  45°  to  H. 

615.  Cylinder  in  Ex.  583.     Draw  the  traces  of  the  tangent  plane  containing 
K(3|  +  l,  x)  on  the  lower  surface. 

616.  Cylinder  in  Ex.  583.     Draw  the  traces  of  the  tangent  plane  containing 
M(4$  +  l-l). 

617.  Cylinder  in  Ex.  583.      Draw  the  traces  of  a  tangent  plane  inclined 
60°  to  H. 

618.  Cylinder  in  Ex.  583.     Draw  the  traces  of  a  tangent  plane  parallel 
toN(l  +  l-l)  O(2,  0-2$). 

619.  Cylinder  in  Ex.  584.     Draw  the  traces  of  the  tangent  plane  through 
P(3$  +  2,  y)  on  the  upper  surface. 

620.  Cylinder  in  Ex.  584.     Draw  the  traces  of  a  tangent  plane  through 


621.  Cylinder    in    Ex.   584.       Draw    the    traces    of    two    tangent    planes 
inclined  75°  to  H. 

622.  Cylinder  in  Ex.  584.     Draw  the  traces  of  a  tangent  plane  parallel  to 
B(l  +  l-l)  0(2,  0-2$). 

623.  Cylinder  in  Ex.  586.     Draw  the  traces  of  the  tangent  plane  containing 
D  (3  +  f,  x)  on  the  rear  surface. 

624.  Cylinder  in  Ex.  586.     Draw  the  traces  of  a  tangent  plane  contain- 
ing E(3$  +  $  -2). 

625.  Cylinder  in  Ex.  586.     Draw  the  traces  of  a  tangent  plane  inclined 
45°  to  H.     Use  the  "Normal  method." 

626.  Cylinder  in  Ex.  586.     Draw  the  traces  of  a  tangent  plane  parallel  to 
F(i  +  i_i)  Q(2,  0  —  2$).     "Normal  method." 

627.  Draw  the  projections  of  a  l$-in.  circular  cylinder  tangent  to  the  plane 
T(l+2$)  4(2$  -H)  along  the  line  A(l+2$,  0)  B(2$,  0-li). 

PRACTICAL  EXERCISES 

Note.  —  Some  of  these  exercises  combine  the  cylinder  and  cone.     When 
pieces  are  duplicated,  develop  only  one. 

628.  Develop  the  patterns  for  the  two-piece  8-in.  pipe  elbow  (Fig.  92). 
Any  convenient  scale. 


CURVED  LINES  AND  SURFACES 


143 


J |_.JL 

k- io'- >l 

3 -Piece  Pipe  Elbow. 


K-— -<? * 

3 -Piece  Oblong  Elbow.  ^x 

SZ-A 

<  /         R.I 
/\  / 


Conical  Reducing  Elbow.  Oblong   Hood. 

FIG.  92. — Practical  examples  of  cylindrical  and  conical  surfaces. 


144  PRACTICAL  DESCRIPTIVE  GEOMETRY 

629.  Develop  the  patterns  for  the   Locomotive   Stack  in  Fig.   92.     Any 
convenient  scale. 

630.  Lay  out  the  patterns  for  the  3-piece  6-in.  pipe  elbow  (Fig.  92).     This 
pipe  is  composed  of  circular  cylinders.     Figure  the  angles  for  the  joints 
so  as  to  keep  the  diameter  uniform. 

631.  Design  a  4-piece  6-in.  pipe  elbow  for  a  90°  turn,  and  lay  out  the  patterns 
to  scale. 

632.  Design  a  4-piece  8-in.  pipe  elbow  for  a  105°  turn,  and  lay  out  the 
patterns  to  scale. 

633.  Design  a  5-piece  6-in.  pipe  elbow  for  a  90°  turn,  and  lay  out  the  patterns 
to  scale. 

634.  Lay  out  the  patterns  for  the  oblong  elbow  (Fig.  92)  to  scale. 

635.  Design  a  4-piece  oblong  elbow,  of  the  same  section  as  in  Fig.  92  (Ex. 
634),  for  a  90°  turn,  and  lay  out  the  patterns. 

636.  Design  a  4-piece  oblong  elbow,  of  the  same  section  as  in  Fig.  92  (Ex. 
634),  for  a  120°  turn,  and  lay  out  the  patterns. 

637.  Lay  out  the  patterns  for  the  piece  of  helical  tubing  (Fig.  92) .     Full  size. 

638.  Lay  out  the  patterns  to  scale  for  the  scale  scoop  (Fig.  92) . 

639.  Lay  out  the  patterns  for  the  conical  reducing  offset  elbow  (Fig.  92) . 
The  4-in.  and  10-in.  pipes  connected  by  the  elbow  are  both  of  circular 
section,  and  the  bases  of  the  cone  are  perpendicular  to  its  axis.     De- 
velop the  three  pieces  to  scale. 

640.  Lay  out  the  patterns  to  scale  for  the  corrugated  galvanized  iron  pipe 
(Fig.  92) .     Design  your  own  sectional  outline. 

641.  Lay  out  the  pattern  to  scale  for  the  oblong  reducing  hood  (Fig.  92). 
Note. — Use  two  spaces  for  all  these  exercises,  except  637. 

THE  CONVOLUTE 

189.  The  Convolute,  the  third  class  of  single-curved  surfaces, 
is  generated  by  the  tangent  line  to  any  space  curve,  as  it  moves 
along  the  curve.     The  tangent  line  in  any  position  contains  two 
consecutive  points  of  the  curve  (see  Definition,  Art.  118).     This 
means  that  any  two  consecutive  tangents  to  any  curve  have  one 
point  in  common;  thus,  A,  B  and  C  are  three  consecutive  points 
in  a  curve,  and  AB  and  BC  are  therefore  two  consecutive  tan- 
gents with  B  as  their  intersection.     Thus  all  consecutive  tan- 
gents to  any  space  curve  intersect,  and  therefore  become  ele- 
ments of  a  single-curved  surface.     The  convolute  is,  therefore,  a 
single-curved  surface. 

190.  Helical  Convolute. — The  most  important  surface  of  this 
class  is  that  generated  by  the  tangent  to  the  helix.     Fig.  93 
shows  both  projections.     It  is  sometimes  called  the  "  Developable 
Helicoid"  and  is  found  in  practical  use  in  some  designs  of  screw 
conveyors. 

The  surface  is  limitless,  the  portion  here  shown  being  only  a 


CURVED  LINES  AND  SURFACES 


145 


"  three-quarter  turn."  In  practical  design  it  is  always  limited 
by  a  cylinder  and  upper  and  lower  bases.  The  //-trace  of  the 
surface  is  an  involute,  and  any  plane  perpendicular  to  the  axis 
will  cut  the  surface  in  an  involute  of  a  circle  whose  diameter  is 
that  of  the  helical  cylinder. 

Note. — In  all  the  problems  relating  to  convolutes  the  axis  will  be  taken 
perpendicular  to  H,  as  in  Fig.  93. 


FIG.  93. — Projections  of  a  convolute. 

191.  Problem  77. — To  locate  a  point  on  the  surface  of  a  con- 
volute, one  projection  being  given. 

First  Case. — When  the  H-projection  is  given. 

Analysis. — 1.  Through  the  given  projection  draw  the  tangent 
to  the  circle,  which  is  the  //-projection  of  the  helix,  intersecting 
the  involute  trace. 

2.  Draw  the  F-projection  of  this  tangent,  by  projecting  the 
point  of  tangency  to  the   F-projection  of  the  helix,   and  the 
involute  intersection  to  GL. 

3.  Having  both  projections  of  the  element  through  the  point, 
locate  on  it  the  F-projection  of  the  point. 

Second  Case. — When  the  V-projection  is  given. 

Analysis. — 1.  Pass  a  plane  through  the  point,  parallel  to  H, 
and  obtain  its  intersection  (involute)  with  the  surface. 

2.  Project  from  the  F-projection  of  the  point  to  the  involute, 
thus  obtained,  to  locate  the  //'-projection. 
10 


146  PRACTICAL  DESCRIPTIVE  GEOMETRY 

192.  Problem  78. — To  pass  a  plane  tangent  to  a  convolute 
through  a  point  on  the  surface. 

1.  Draw  the  element  through  the  point.     This  is  one  line  of 
the  required  tangent  plane. 

2.  Where  the  element  of  contact  intersects  the  involute,  draw 
the  tangent.    %This  will  be  the  #-trace  of  the  tangent  plane. 

3.  Find  the  F-trace  by  any  of  the  known  methods. 

193.  Problem  79. — To  pass  a  plane  tangent  to  a  convolute 
through  a  point  outside  the  surface. 

Analysis. — 1.  With  the  given  point  as  apex,  draw  a  right 
circular  cone,  its  base  on  H,  its  elements  having  the  same  incli- 
nation to  H  as  those  of  the  convolute. 

2.  Pass  a  plane  tangent  to  both  cone  and  convolute.  This 
will  be  the  required  tangent  plane. 

Second  Method.— Analysis. 

1.  Pass  a  plane  parallel  to  H  through  the  given  point. 

2.  Obtain  its  intersection  with  the  convolute. 

3.  Draw  the  tangent  from  the  given  point  to  this  intersection. 

4.  This  tangent  and  the  element  of  contact  will  determine  the 
tangent  plane. 

194.  Problem  80. — To  find  the  intersection  of  a  convolute  by 
a  plane. 

Analysis. — Find  the  piercing  points  of  a  sufficient  number  of 
elements  with  the  given  plane  and  draw  a  smooth  curve  through 
these  points. 

Note. — As  the  surface  ends  at  the  directrix  curve  (practically,  if  not  theo- 
retically), any  piercing  points  that  lie  beyond  the  points  of  tangency  are 
not  considered.  A  plane  that  contains  an  element  will  cut  from  the  surface 
a  straight  line  until  it  passes  the  point  of  tangency  of  that  element,  when  it 
will  begin  cutting  a  curve. 

195.  Problem  81. — To  develop  the  surface  of  a  convolute. 
Discussion. — If  the  helical  directrix  be  rolled  into  a  plane, 

element  by  element,  it  will  be  seen  to  have  a  constant  radius  of 
curvature,  and  will  therefore  develop  as  a  circle.  This  radius 
is  the  normal  line  at  any  point  of,  the  helix,  that  intersects  the 
axis,  because  all  such  normals  will  have  equal  lengths.  The 
elements  of  the  convolute,  being  tangent  to  the  helix,  are  per- 
pendicular to  the  respective  normals,  and  when  the  rolling  out 
occurs,  these  elements  will  be  consequently  tangent  to  the  circle 
made  by  developing  the  helix.  As  the  lengths  of  the  tangents 


CURVED  LINES  AND  SURFACES 


147 


are  proportional  to  their  ^-projections,  the  developed  curve  of 
their  extremities  will  be  the  involute  to  the  developed  helix. 
This  is  so,  because  each  tangent  is  equal  in  length  to  the  helix 
from  the  origin  to  the  point  of  tangency,  and  when  helix  is 
developed  into  a  circle,  the  tangents  are  equal  to  the  arc  from 
the  origin  to  the  point  of  tangency.  This  corresponds  to  the 
generation  of  the  involute. 

Analysis. — The  entire  analysis  is  practically  confined  to  the 
determination  of  the  radius  of  the  circle  of  the  developed  helix. 

Graphical  Solution. — Take  the  tangent  BF  which  is  parallel  to 
V,  Fig.  94.  At  the  point  d' ,  where  b'f  crosses  the  F-projection 


FIG.  94. — The  convolute  and  its  development. 


of  the  outside  element  of  the  cylinder,  draw  c'd'  perpendicular 
to  f'b',  and  at  b'  draw  a  horizontal  line  b'c'  intersecting  the 
perpendicular  c'd'  at  c'.     The  required  radius  is  then  c'b'. 
Construction.  —  1.   Draw  the  circle  of  radius  equal  to  c'b'. 

2.  Lay  off  arcs  aj^,  a^^a^,  etc.,-  making  them  equal  to 
the  true  lengths  of  the  corresponding  tangents  to  the  helix. 

3.  Draw  the  involute  from  at  to  r1;  and  the  plane  figure  included 
within  the  involute  is  the  development. 

Mathematical  Value.  —  The  mathematical  value  for  the  radius 


r> 


of  the  development  is  the  expression  -  — 


in  which  R  is  the 


radius  of  the  helical  cylinder,  and  0  is  the  inclination  of  the 
elements  to  the  horizontal. 


148  PRACTICAL  DESCRIPTIVE  GEOMETRY 


196.  EXERCISES 

642.  Draw  a  270°   convolute    to    the    helix,  whose   axis   is   A(l£,  0-1{) 
B(l^  +  2  — 1|),    1-in.   diameter,  2-in.  pitch.     Find   the  traces   of   the 
tangent  plane  containing  C(2J,  x  — 2)  on  the  surface. 

643.  Draw  the  convolute  in  Ex.  642.     Draw  the  traces  of  the  tangent  plane 
through  D(lf  +  i,  y)  on  the  front  surface. 

644.  Draw  a  270°  convolute  to  the  "left-hand"  helix,  whose  axis  is  E(4^, 
0-1)  F(4i  +  3-l),  11-in.  diameter,  2£-m.  pitch.     Draw  the  traces  of 
a  plane  tangent  at  the  point  G(4|,  x—  If)  on  the  surface. 

645.  Draw  the  convolute  in  Ex.  644.     Draw  the  traces  of  the  plane  tangent 
at  K(3|  +  l,  y)  on  the  front  surface. 

646.  Draw  the  convolute  in  Ex.  642.     Draw  the  traces  of  a  plane  tangent 
containing  M(l+2-2i). 

647.  Convolute    in    Ex.    644.     Draw   the    traces    of    the    tangent    plane 
containing  N(4i  +  l-2|). 

648.  Convolute  in  Ex.  642.     Find  its  intersection  with  the  plane  S(l  +  l) 
4(1-3). 

649.  Convolute  in  Ex.  644.     Find  its  intersection  with  the  plane  T(  +  1J) 
oo(-3). 

650.  Develop  the  surface  of  the  convolute  in  Ex.  642. 

651.  Develop  the  surface  of  the  convolute  in  Ex.  644. 

652.  Find  the  shortest  path  on  the  surface  of  the  convolute  in  Ex.  642  from 
O(lf,  x-2)  toP(3,  y-1). 

653.  Find  the  shortest  path  on  the  surface  of  the  convolute  in  Ex.   644 
from  C(5,  x-lf)  to  D(3,  y-£). 

197.  General  Notes  and  Rules  on  Single-Curved  Surfaces. 

1.  A  tangent  plane  to  any  single-curved  surface  is  tangent  at 
every  point  on  the  element. 

2.  A  tangent  plane  to  a  cylinder  is  always  parallel  to  the  axis, 
and  to  all  elements. 

3.  A  tangent  plane  to  a  cone  must  contain  the  apex,  and  inter- 
sects every  element  but  the  element  of  contact. 

4.  To  pass  a  plane  tangent  to  a  cylinder,  draw  a  tangent  line 
and  a  line  intersecting  the  tangent,  parallel  to  the  axis.     These 
two  lines  determine  the  required  plane. 

5.  To  pass  a  plane  tangent  to  a  cone :  a  tangent  line  and  the 
apex  are  sufficient  to  determine  it. 

6.  The  element  of  contact  and  a  tangent  line  at  any  point  of  it 
determine  the  tangent  plane  to  any  single-curved  surface. 

7.  Tangent  planes  to  certain  single-curved  surfaces  will  cut 
the  surface,  but  not  at  the  point  of  tangency. 

8.  To  cut  an  element  from  a  cone,  pass  a  plane  through  the 
apex. 


CURVED  LINES  AND  SURFACES  149 

9.  To   cut  an  element  from  a  cylinder  pass  a  plane  parallel 
to  the  axis. 

10.  To  cut  an  element  from  a  helical  convolute,  pass  a  plane 
tangent  to  the  cylinder  of  the  helix. 

11.  In  the  development  of  a  cone,  all  the  elements  radiate  from 
the  apex.     Only  a  right  circular  cone  will  roll  out  in  a  segment  of 
a  circle.     All  other  cones  must  be  developed  by  triangulation. 

12.  In   the  development  of  a  cylinder,  all  the  elements  roll 
out  parallel.     Only  a  right  section  will  roll  out  in  a  straight  line. 

WARPED  SURFACES 

198.  Discussion. — According  to  the  definition  in  Art.  152,  a 
warped  surface  is  generated  by  moving  a  straight  line  so  that  no 
two  consecutive  positions  shall  be  in  the  same  plane.     It  thus 
falls  into  the  classification  with  single-curved  surfaces  as  a  ruled 
surface.     Planes  may  be  passed  through  any  point  of  a  warped 
surface  that  will  cut  straight  lines  from  the  surface,  always  one, 
and,   in  the   case   of  two   surfaces,   two   straight  lines.     Other 
planes  cut  curves  from  them. 

Warped  surfaces  are  of  many  varieties  and  are  common  in 
practical  work.  In  many  cases,  when  they  occur  in  practical 
work,  it  is  not  necessary  for  the  draftsman  to  recognize  their 
peculiarities,  but  very  often,  as  in  arches,  sheet  metal  work, 
propeller  blades,  skew  gears,  conveyors,  etc.,  it  is  highly  impor- 
tant that  engineers  and  draftsmen  know  their  properties  and 
methods  of  generation. 

199.  Methods     of    Generation. — Warped     surfaces     may    be 
generated: 

1.  By  revolving  one  line  about  another  not  in  the  same  plane. 

2.  By  moving  a  straight  line,  the  generatrix,  along  two  lines, 
directrices,  either  straight,  curved, or   both,  keeping  parallel  to 
some  plane,  called  the  Plane  Director. 

3.  By  moving  a  straight  line  along  three  directrices,  straight, 
curved,  or  both. 

4.  By  moving  a  straight  line  along  an  axis   at  a  constant 
angle  with  the  axis,  following  a  space  curve  as  a  directrix. 

Other  methods  might  be  invented,  but  these  cover  all  the 
cases  met  in  practical  work. 

200.  Tangent  Planes. — In  Art.  155  we  find  the  general  rule 
for  tangent  planes  to  all  Ruled  Surfaces,  as  follows :  One  tangent 


150  PRACTICAL  DESCRIPTIVE  GEOMETRY 

line  and  the  element  at  any  point  determine  the  tangent  plane  at 
that  point. 

In  general,  a  tangent  plane  is  only  tangent  at  one  point  of  a 
warped  surface.  There  are  certain  points,  as  in  conoids  and 
warped  cones,  where  the  tangent  plane  will  be  tangent  all  along 
the  element,  but  such  exceptions  will  be  readily  recognized. 

All  planes  tangent  to  warped  surfaces  (except  in  the  cases  just 
noted)  intersect  the  surface,  because  they  are  not  tangent  all 
along  the  element,  and,  conversely,  any  plane  that  intersects  a 
warped  surface  in  an  element  will,  in  general,  be  tangent  to  the 
surface  at  some  point. 

201.  Double-Ruled  Surfaces. — Two  surfaces,  the  Hyperboloid 
of  Revolution  of  One  Nappe,  and  the  Hyperbolic   Paraboloid,  are 
called  double  ruled  surfaces  from  the  fact,  as  will  be  shown  later, 
that  they  are  capable  of  generation  by  two  different  straight 
lines.     This  yields  the  phenomenon,  that  through  every  point 
in  the  surface  two  straight  lines  (elements)  can  be  drawn  lying 
wholly  within  the  surface.     From  this  fact  we  have  the  special 
rule  for  tangent  planes  to  these  surfaces;  The  two  elements  of  a 
double -ruled  surface  through  any  point  of  the  surface  determine 
the  tangent  plane  at  that  point. 

202.  Representation. — Some  warped  surfaces  can  be  shown 
in  outline  in  orthographic  projection  with  sufficient  clarity,  and 
others  must  be  shown  by  drawing  a  sufficient  number  of  elements. 
In  practical  drafting,  where  the  surface  is  always  limited,  the 
outlines  are  usually  sufficient. 

203.  Location  of  Points. — To  locate  a  point  on  any  warped 
surface,  one  projection  being  given,  find  the  piercing  point  of 
the  projector    of  the  point  with  the  surface.     This  is  done  by 
passing  a  plane  through  the  projector,   and  finding  its  inter- 
section with  the  surface.     Where  the  projector  intersects  this 
line  will  be  the  location  of  the  point.     In  many  cases  it  will  be 
easy  to  draw  the  element  through  the  point,  simplifying  the 
operation. ' 

204.  Developments. — Strictly    speaking,    a    warped    surface 
cannot  be  developed,  because  no  two  consecutive  elements  lie  in 
the  same  plane,  and  cannot,  therefore,  be  rolled  into  a  plane 
without  " warping"  them;  that  is,  destroying  the  relation  be- 
tween them,     However,   there   are   approximate   developments 
that  give  very  satisfactory  results,  almost  as  good  us  the  devel- 
opment of  a  cone  by  triangulation.     These  warped  develop- 


CURVED  LINES  AND  SURFACES 


151 


ments  are  obtained  also  by  triangulation.  A  sufficient  number 
of  elements  are  drawn,  dividing  the  surface  into  minute  warped 
quadrilaterals,  in  which  the  amount  of  warp  is  so  slight  as  to  be 
negligible.  Diagonals  are  then  drawn  across  each  of  these 
quadrilaterals,  dividing  them  into  triangles.  These  triangles 
are  then  measured  and  placed  in  order  in  their  (approximate) 
true  size.  The  resulting  surface,  obtained  by  bending  back  this 
development,  cannot  be  the  exact  surface,  as  shown  in  the  draw- 
ing, but  will  be  near  enough  for  practical  purposes.  In  making 
single-curved  surfaces  from  flat  patterns  there  need  be  no  dis- 
tortion of  the  metal,  but  in  making  up  a  warped  surface,  there 
will  be  some  distortion. 

THE  HELICOID 

205.  The  most  frequent  warped  surface  encountered  in  draft- 
ing, is  the  Helicoid.  It  is  generated  by  Method  4,  Article  199, 
by  moving  a  straigfit  line  along  a  helix  and  its  axis,  keeping  a 
constant  angle  with  the  axis.  The 
helix  and  its  axis  are  the  directrices. 
The  surface  is  theoretically  unlimited, 
but  in  practise  is  usually  included 
between  two  helices.  Any  circular 
cylinder,  having  its  axis  coincident 
with  that  of  the  helical  directrix,  will 
intersect  the  helicoid  in  a  helix.  Any 
plane  perpendicular  to  the  axis  will 
cut  the  helicoid  in  a  spiral  of  Arch- 
imedes. Any  plane  passed"  through 
the  axis  will  intersect  the  helicoid  in 
an  element. 

Fig.  95  shows  a  heliQoid,  with  its 
axial  angle  30°.  In  drawing  the  helix 
ascertain  number  of  points  were  laid 
out,  and  through  these  points  lines 
intersecting  the  axis  are  drawn,  mak- 
ing the  given  angle  with  the  axis.  As  the  helix  is  everywhere 
equally  distant  from  the  axis,  this  is  simple;  as  follows. 

1.  Draw  one  element  parallel  to  V,  which  will  be  drawn  in  its 
true  angle  with  the  axis. 

2.  As  the  generatrix  is  moved,  it  climbs  both  axis  and  helix 
in  exactly  the  same  upward  distances.     Therefore,  fronr  any 


FIG.  95.— Helicoid. 


152  PRACTICAL  DESCRIPTIVE  GEOMETRY 

point  on  the  helix,  an  element  may  be  drawn  to  a  point  on  the 
axis  that  is  the  same  distance  up,  measured  on  the  F-projec- 
tion.  Thus  in  Fig.  95,  the  point  b'  is  the  same  distance  above 
a'  as  the  point  2  is  above  1  on  the  axis. 

The  point  P  is  located  on  the  surface  by  drawing  the  element 
through  the  point  and  locating  the  other  projection  on  the 
element.  If  the  F-projection  of  the  point  is .  given,  pass  a 
plane  through  the  point  parallel  to  H.  This  will  cut  the  ele- 
ments in  a  curve  on  which  -the  //"-projection  will  be  located. 

The  helicoid  is  the  surface  of  the  V-thread,  the  square  thread, 
spiral  springs,  screw  conveyors,  certain  fire-escapes,  the  twist 
drill,  the  spiral  runway,  helicoidal  arch,  and  numerous  other 
surfaces  frequent  in  machine  and  structural  design. 

Helicoidal  arch. — In  the  design  of  skew  arches,  that  is,  arches 
with  their  axes  oblique  to  one  or  both  of  the  openings,  warped 
surfaces  are  often  encountered,  cylindroids  or  cow's  horns 
being  the  usual  surfaces  in  such  arches.  The  arch  called  the 
helicoidal  arch  takes  its  name  not  from  the  character  of  its 
surface,  as  might  be  supposed,  but  from  the  surface  of  the 
"  coursing  joints."  The  line  of  the  joint  is  laid  out  as  a  straight 
line  on  the  developed  surface  of  the  arch,  and,  when  brought  into 
its  actual  position,  it  becomes  a  helix.  The  surface  of  the  joint 
is  generated  by  a  line  normal  to  the  surface  of  the  arch  moving 
along  the  helix,  radial  to  the  axis  of  the  helix.  This  surface  is 
therefore  identified  as  a  helicoid.  Inasmuch  as  the  developed 
helix  is  a  straight  line  the  work  of  laying  out  the  blocks  is  very 
simple.  The  seam,  being  helicoidal,  must  be  laid  out  with  care, 
to  ensure  good  fitting  blocks. 

THE  HYPERBOLOID  OF  REVOLUTION  OF  ONE  NAPPE 

206.  This  peculiar  surface  is  the  only  warped  surface  that  is 
a  surface  of  revolution.  It  is  generated  by  Method  1,  Article  199, 
by  revolving  a  straight  line  about  an  axis  not  in  the  same  plane. 
Fig.  96  shows  this  surface  represented  by  its  elements.  Either 
projection  of  all  the  elements  could  be  inverted,  and  there  would 
be  no  change  in  the  surface.  If  AB  be  considered  an  element, 
its  F-projection  might  be  either  of  the  lines  marked  a'b',  and 
the  same  surface  would  be  generated  by  revolving  either  line 
about  the  axis.  This  makes  it  a  double-ruled  surface,  and  a 
tangent  plane  is  determined  at  any  point  by  two  elements 
through  that  point. 


CURVED  LINE  AND  SURFACE 


153 


^  207.  To  draw  the  Elements. — All  points  in  the  generatrix 
move  in  circles,  hence  there  is  a  vertex  to  the  curve  of  the  outline, 
a  smallest  circle,  called  the  Gorge  Circle,  generated  by  the  point 
nearest  the  axis.  All  elements  intersect  the  gorge  circle  and 
their  //-projections  are  all  tangent  to  its  //-projection.  Two 
limiting  planes,  bases,  are  always  necessary,  always  taken  per- 
pendicular to  the  axis,  and  usually  at  equal  distances  from  the 
gorge  circle.  Being  perpendicular  to  the  axis,  these  bases  are 
circular.  Why?  One  base  is  usually  in  H,  and  the  elements 


FIG.  96. — Hyperboloid  of  revolution. 

terminate  in  the  bases.  Hence,  1.  Draw  the  gorge  and  base 
circles.  2.  Draw  as  many  elements  as  desirable,  16  or  24,  the 
//-projections  tangent  to  the  gorge  circle  and  terminating  in 
the  base  circles.  3.  Project  the  terminals  of  the  elements  to 
the  y-traces  of  the  respective  base  planes. 

Note. — When  the  bases  are  symmetrically  placed,  there  will  be  only  one 
circle  projected  on  H,  as  in  Fig.  96. 

208.  To  Locate  a  Point  on  the  Surface. — If  the  given  projec- 
tion is  on  //,  draw  an  element  through  the  given  projection  and 
project  the  point  to  the  7-projection  of  the  element. 


154  PRACTICAL  DESCRIPTIVE  GEOMETRY 

If  the  given  projection  is  on  V,  pass  a  plane  through  the  sur- 
face, parallel  to  H,  through  the  given  projection.  This  will  cut 
a  circle  from  the  surface,  projected  in  its  true  size  on  H.  The 
//-projection  will  be  in  the  circle. 

209.  To  Pass  a  Plane  Tangent  at  any  Point  on  the  Surface.— 
Draw  the  two  elements  through  the  given  point.  They  will 
determine  the  tangent  plane. 

Tangent  planes  to  this  surface,  as  to  all  surfaces  of  revolution, 
are  perpendicular  to  the  plane  of  the  axis  and  the  point  of  tan- 
gency,  called  the  Meridian  Plane. 


FIG.  97. — Rolling  hyperboloids.  (Both  surfaces  are  generated  by  revolving  the  same 
straight  line  about  two  different  axes,  and  they  thus  have  one  element  in  common,  and 
are  tangent  to  each  other  along  this  element.) 

The  Meridian  Section  of  this  surface  is  a  hyperbola,  and  the 
same  surface  can  be  generated  by  revolving  a  hyperbola  about 
its  vertical  axis.  If  proof  of  this  is  desired,  it  can  be  found  in 
Church  and  Bartlett,  page  150. 

The  most  important  function  of  the  hyperboloid  is  found  in 
the  pitch  surfaces  of  skew  gears.  If  one  line  be  successively 
revolved  about  two  axes,  not  in  its  plane,  the  line  will  be  simul- 
taneously an  element  of  both  hyperboloids;  that  is,  the  two 
hyperboloids  generated  will  be  tangent.  These,  if  used  as  the 
basis  of  gears,  form  an  ideal  transmission  medium  through 
shafts  not  in  the  same  plane. 

THE  HYPERBOLIC  PARABOLOID 

210.  This  surface  is  generated  by  moving  a  straight  line  along 
two  straight  lines,  not  in  the  same  plane,  keeping  it  always 


CURVED  LINES  AND  SURFACES 


155 


parallel  to  a  plane  director.  The  elements  are  all  parallel  to 
this  plane,  but  not  to  each  other. 

Representation. — If  a  definite  position  of  the  two  directrices  is 
given,  the  elements  may  be  drawn  without  the  assistance  of 
the  plane  director,  by  dividing  each  directrix  into  the  same 
number  of  equal  parts  and  connecting  the  points  in  order.  This 
is  so,  because  a  series  of  parallel  planes  will  cut  any  two  lines 
in  proportional  segments. 

211.  Second  Generation. — If  any  two  elements  be  taken,  as 
BD  and  AC  (Fig.  98)  and  divided  as  were  the  directrices  AB  and 
CD,  and  joined  in  order,  the  surface  thus  generated  would  be 


FIG.  98. — Hyperbolic  paraboloid,  showing  double  ruling. 


identical  with  the  first  surface.  In  the  two  generations,  the 
directrices  of  the  first  become  elements  of  the  second,  and  the 
elements  of  the  first  become  directrices.  This  makes  it  a  double- 
ruled  surface,  and  the  rule  for  tangents  is  the  same  as  for  the 
hyperboloid. 

212.  It  is  named  the  Hyperbolic  Paraboloid  because  certain 
planes  cut  hyperbolas  from  the  surface,  others  parabolas,  while 
tangent  planes  cut  straight  lines.  It  is  chiefly  of  interest  to 
the  mathematician,  but  it  is  occasionally  met  in  practical  work, 
and  is  then  usually  called  a  "  Warped  Quadrilateral." 


156  PRACTICAL  DESCRIPTIVE  GEOMETRY 

THE  CYLINDROID 

213.  The  Cylindroid  is  a  warped  surface  with  two  curved-line 
directrices   and  a   plane   director.     It   differs   from   a   cylinder 
in  that  its  elements,  while  parallel  to  one  plane  are  not  parallel 
to  each  other.      It  is  found  in   practical  work   in  arches  and 
sheet  metalwork. 

THE  CONOID 

214.  The  Conoid  has  one  curved-line  directrix,  one  straight- 
line  directrix,  and  a  plane  director. 

A  helicoid,  whose  elements  are  at  right  angles  to  the  axis, 
may  be  regarded  as  a  conoid. 

THE  COW'S  HORN 

215.  This  surface  has  three  directrices,  two  being  circles,  and 
the  third  a  straight  line.     The  circles  are  usually  in  parallel 
planes  with  their  centers  in  a  plane  perpendicular  to  these  par- 
allel planes.     The  straight  line  directrix  usually  lies  in  the  plane 
of  the  centers.     The  Cow's  Horn  is  frequently  encountered  in 
stone  arch  designs,  and  occasionally  in  sheet  metal  hoods. 

THE  WARPED  CONE 

216.  This  surface,  often  found  in  reducing  hoods  and  pipe 
connections,   has   three   directrices;   as   follows,   two   circles   or 
ellipses   (or  one  of  each)   in  planes  not    parallel,  and    the  line 
joining  their  centers,  called  the  axis. 

217.  Other   Warped  Surfaces. — Besides   the  foregoing,   there 
are   many  forms,   which  have  no   specific  names.     They  may 
have   three    straight    directrices,    three    curved    directrices,    or 
combinations   of   these  with  plane   directors.     Whenever  they 
are  met,  they  will  present  no  difficulties  to  the  student,  who  is 
familiar  with  those  described. 

218.  The  Development  of  a  Specimen  Warped  Surf  ace. — Develop 
the  helicoid  shown  in  Fig.  99,  for  a  half  turn,  between  H  and  the 
helical  directrix. 

1.  Draw  the  elements  Al,  B2,  etc. 

2.  Draw  diagonals  A2,  B3,  etc.,  of  the  quadrilaterals. 

3.  Measure   the   true   lengths   of   the   various   elements   and 
diagonals.     The  construction  on  the  left  shows  a  simple  arrange- 
ment for  accomplishing  this.     Lay  off  the  heights  of  the  various 
points,  a',  b' ,  etc.,  on  a  vertical  line,  and  lay  off  the  lengths  of 


CURVED  LINES  AND  SURFACES 


157 


the  various  H-projections  on  a  horizontal  line,  elements  on  one 
side,  diagonals  on  the  other.  Their  true  lengths  will  be  the 
hypothenuses,  a'l  or  a'2,  etc. 


lO                   n 

-~~"          \\           ^\ 

V      \\     /fXj 

t 

XN\     \  \    /  \    i 

\  \    ,  \   i  •  \ 

\ 

\  \  .'  \  i 

\     Nx        \       \    \        \      1 

\      \     1       \  I       \  / 

^,     \    \  \     \'     './ 

xx 

4.  Rectify  the  various  arcs  of  the  helix  and  spiral.     The  arcs 
of  the  helix  are  equal  in  length  to  corresponding  arcs  of  a  circle 


,  ,.  OP 

whose  radius  =—   — . 
cos26 


(See  the  graphical  solution  given  in  Art. 


158  PRACTICAL  DESCRIPTIVE  GEOMETRY 

196.)  The  lengths  of  the  spiral  arcs  are  obtained  by  stepping  off 
small  chords  with  the  dividers. 

5.  With  these  materials,  construct  the  triangles  in  order, 
as  shown  on  the  right. 

This  method,  with  such  modifications  as  may  be  necessary, 
is  employed  in  the  development  of  all  warped  surfaces.  It  is 
laborious,  but  there  is  no  shorter  method.  The  important 
operation  in  this,  as  in  all  problems  relating  to  surfaces,  is  to 
get  the  elements. 

219.  EXERCISES  IN  WARPED  SURFACES 

654.  Draw  the  projections  of  a  right-hand  helicoid,  its  axis  A(2,  0—1$) 
B(2  +  3  —  1$),  and  its  helical  directrix  of  l$-in.  diameter  and  2$-in.  pitch. 
The  constant  angle  between  elements  and  axis  is  45°.     Draw  the  H- 
trace  of  the  helicoid,  and  locate  C(2±  +  l,  x)  and  D(3,  y-2|)  on  the 
surface. 

655.  Draw  the  projections  of  a  left-hand  helicoid,  its  axial  directrix  being 
M(4,  0  — 1|)  N(4  +  3  — lj),  its  helical  directrix  having  a  lj-in.  diameter 
and  2-in.  pitch.     The  constant  angle  between  elements  and  axis  is 
60°.     Draw  the  H-tr&ce  of  the  helicoid,  and  locate  O(4|  +  |,  x)  and 
P(3,  y  —  2)  on  the  surface. 

656.  Helicoid  in  Ex.  654.     Draw  the  traces  of  the  tangent  plane  at  C(2£  + 
1,  x)  or  D(3,  y  —  2$)  on  the  surface. 

657.  Helicoid  in  Ex.  655.     Draw  the  traces  of  the  tangent  plane  at  O(4f  + 
|,  x)  or  P(3,  y  — 2)  on  the  surface. 

658.  Helicoid  in  Ex.  654.     Find  the  points  in  which  it  is  pierced  by  the  line 
E(l+f-l)F(5  +  f-3). 

659.  Helicoid  in  Ex.  655.     Find  the  points  in  which  it  is  pierced  by  the 
line  K(2  +  l$,  0)L(5$,  0-3). 

660.  Helicoid  in  Ex.  654.     Find  its  intersection  with  the  plane  Q(3  +  3) 
3(3-3). 

661.  Helicoid  in  Ex.  654.     Find  its  intersection  with  the  plane  R(l  +  l) 
5(1-3).     - 

662.  Helicoid  in  Ex.  655.     Find  its  intersection  with  the  plane  S,  parallel 
to  H,  |  in.  above  H. 

663.  Helicoid  in  Ex.  655.     Find  its  intersection  with  the  plane  T(2  +  2) 
2(5-3|). 

664.  Draw  the  projections  (24  elements)  of  a  hyperboloid  of  revolution,  its 
axis  being  A(3,  0-1$)  B(3  +  3-l$)  and  one  element  being  C(2  +  2$- 
f)  D(4,  0  — f).     Locate  E(2$  +  $,  x)  on  the  rear  surface,  and  draw  the 
traces  of  the  plane  tangent  at  E. 

665.  Draw  the  projections  (24  elements)  of  a  hyperboloid  of  revolution, 
axis  K(3  +  H,  0)  L(3  +  l|-3),  one  element  M(2  +  f-2$)  N(4  +  f,  0). 
Locate  O(3$,  y  — 1)  on  the  under  surface,  and  draw  the  traces  of  the 
tangent  plane  at  O. 

666.  Hyperboloid  in  Ex.  664.     Find  its  intersection  with  the  plane  Q(3$  +  3) 
3$  (3$ -3).     Show  profile  view. 


CURVED  LINES  AND  SURFACES  159 

667.  Hyperboloid  in  Ex.  665.     Find  its  intersection  with  the  plane  R (2 +  3) 
2(5  —  2).     Show  its  true  size. 

668.  Hyperboloid  in  Ex.  664.     Find  its  intersection  with  the  plane  S(  +  l) 
oo(  — 3).     Show  its  true  size. 

669.  Hyperboloid  in  Ex.  665.     Find  its  intersection  with  the  plane  T(2  +  3) 
5(2  —  2).     Show  its  true  size. 

670.  Hyperboloid  in  Ex.  664.     Find  the  points  where  it  is  pierced  by  the 
line  F(l +  1-1)  G(5  +  l-3). 

671.  Hyperboloid  in  Ex.  665.     Find  the  points  in  which  it  is  pierced  by 
A(l  +  3,  0)  B(5,  0-2). 

672.  Hyperboloid  in  Ex.  664.     Do  not  draw  the  elements.     Pass  a  plane 
through  the  line  F(2  +  l,  0)  G(5  +  l-2)  that  will  be  tangent  to  the 
hyperboloid  and  find  the  point  of  tangency.     (Hint:   A  plane  through 
the  axis,  perpendicular  to  the  plane  through  FG,  will  contain  the  point 
of  tangency,  Art.  209.     Also  see  Art.  200.) 

673.  Hyperboloid  in  Ex.  665.     Pass  a  plane  through  the  line  A(l  +  3,  0) 
B(4,  0  —  2)  that  will  be  tangent  to  the  hyperboloid  and  find  its  point  of 
tangency.     (Hint:   See  Art.  208  and  209.) 

674.  Let  A(2,  0-2)  B(2  +  3-2)  and  C(i  +  l-l)  D(3$  +  l-l)  be  the  axes 
of  two  hyperboloids  having  a  common  element  E(2f +  2  — If)  F(1J, 
0  — If).     Draw  the  projections,  24  elements,  of  each  hyperboloid,  and 
specify  the  faster  revolving  in  transmitting  power. 

675.  A(i  +  3-li)  B(lf  +  l-i)  and  C(2f  +  l-2$)  D(4J  +  3-l)  are  direc- 
trices of  a  hyperbolic  paraboloid,  and  T(4  +  l)  4(5  — J)  is  its  plane 
director.     Draw  ten  elements  of  the  surface  and  locate  a  point  E 
(2  +  x  — y)  on  the  surface  between  elements  3  and  4. 

676.  K(i  +  l-2)  L(2  +  3-i)  and  M(3£  +  2£-2i)  N(4f  +  l-l)  are  direc- 
trices of  a  hyperbolic  paraboloid,  and  KM  and  LN  are  elements. 
Draw  eight  elements,  and  find  the  traces  of  a  plane  tangent  at  O(2 
+  2^,  x)  on  the  surface. 

677.  A(i  +  3-li)  B(lf-fl-i)  and  C(2f  +  l-2|)  D(4£  +  3-l)  are  direc- 
trices of  a  hyperbolic  paraboloid,  of  which  AC  and  BD  are  elements. 
Find  the  traces  of  a  tangent  plane  at  the  point  E(2  +  x  — y)  on  the 
element  midway  between  AC  and  BD.     Do  not  try  to  show  the  surface 
by  elements. 

678.  A(2,  0,  0)  and  B(3J,  0-2)  are  the  centers  of  two  3-in.  semicircles, 
lying  respectively  in  V  and  T(5  +  2)  5(3$  — 2).     They  are  directrices 
of  a  cylindroid  having  H  as  its  plane  director.     Show  the  surface  by 
12  elements,  and  find  the  traces  of  a  plane  tangent  at  C(3,  x  —  $)  on  the 
surface. 

679.  A  cow's  horn  has  for  directrices  a  3-in.  semicircle  in  V,  center  at  D 
(2,  0,  0),  and  a  li-in.  semicircle  parallel  to  V,  center  at  E(3,  0-f), 
and  the  line  M(2J,  0,  0)  N(2£,  0-3).     Show  the  surface  by  12  elements, 
and  draw  the  traces  of  a  plane  tangent  at  F(3  +  x  — f)  on  the  surface. 

680.  Cow's  horn  in  Ex.  679.     Locate  a  point  on  the  surface  1  in.  above  H 
and  i  in.  in  front  of  V.     Draw  the  traces  of  a  plane  tangent  at  the 
point. 

681.  The  directrices  of  a  conoid  are  a  2-in.  circle  in  H,  center  at  M(2,  0  —  2) 
and  the  line  N(2£  +  2-2)  P(5  +  2-2).     The  plane  director  is  perpen- 


160  PRACTICAL  DESCRIPTIVE  GEOMETRY 

dicular  to  V  and  inclined  60°  to  H.  Show  the  surface  by  24  elements, 
and  find  the  traces  of  a  plane  tangent  at  a  point  on  the  surface  1J  in. 
from  V  and  ^  in.  above  H . 

682.  Conoid  in  Ex.   681.     Intersect  it  by  the  plane  T(1  +  H)    5(2-3J). 
Show  the  true  size  of  the  intersection. 

683.  Develop  the  surface  by  triangulation  of  the  conoid  in  Ex.  682,  between 
H  and  T.     (682  and  683  should  be  given  together  with  one  space  for 
each.) 

684.  A  warped  cone  has  one  base  a  2-in.  circle  in  H,  center  at  A(2,  0  — 1£), 
and  one  base  an  ellipse  in  T(l  +  2)  4(4  +  3)  whose  /^-projection  is  a 
1-in.  circle,  center  at  A.     The  axial  directrix  is  the  line  joining  the 
centers  of  the  circle  and  ellipse.     Show  its  surface  by  16  elements,  and 
find  the  traces  of  a  plane  tangent  at  B(x  +  f  —  1)  on  the  surface. 

685.  Develop  the  warped  cone  in  Ex.  684.     (When  this  exercise  is  given, 
684  and  685  should  be  given  together,  one  space  for  each.) 

686.  A(4  +  2-2)  B(l|  +  |-i),C(2  +  li-i)D(3  +  i-2),  and  E(3i  +  i~3) 
F(4  +  2£  —  J)  are  the  directrices  of  a  warped  surface.    Draw  the  element 
through  B. 

687.  In  the  warped  surface  in  Ex.  686,  find  the  projections  of  a  point  1  in. 
above  H  and  1  in.  from  V. 

PRACTICAL  EXERCISES  IN  WARPED  SURFACES 

In  all  the  development  exercises  use  two  spaces. 

688.  Draw  the  projections  of  a  helical  spring  made  of  £-in.  square  steel, 
outside  diameter  3  in.,  pitch  1^  in.,  2  coils. 

689.  Draw  the  projections  of  a  left  handed  square  thread,  2-in.  diameter, 
f-in.  pitch  (single  thread),  and  3  in.  long. 

690.  Draw  the  projections  of  a  helical  spring  made  of  i-in.  wire,  outside 
diameter  2£  in.,  pitch  1  in.,  and  3  coils. 

691.  Draw  the  projections   of  a   V-thread,    2£-in.   diameter,    ^-in.    pitch, 
3  in.  long.     Show  a  section  made  by  a  plane  perpendicular  to  the  axis. 

692.  Draw  the  projections  of  a  twist  drill,  1-in.  diameter,  5|  in.  long,  made 
from  a  flat  bar  £  in.  thick.     Pitch  of  helix  3£  in.     Sharpen  point  at 
30°.     See  Fig.  100.     Do  not  use  a  ground  line. 

693.  Design  a  screw  conveyor  to  scale,  12-in.  diameter,  capable  of  delivering 
6000  cu.  ft.  of  grain  per  hour,  making  100  revolutions  per  minute. 
Make  the  diameter  of  the  shaft  2  in.     See  Fig.  100. 

694.  Draw  a  conical-helical  spring,  £-in.  round  wire,  large  diameter  2^  in., 
small  diameter  1£  in.,  3  turns,  1-in.  pitch. 

695.  Draw  the  projections  of  the  pitch  surfaces  of  two  hyperboloidal  gears, 
operating  at  45°,  having  a  gorge  circle  ratio  of  5  to  3. 

Note. — Draw  one  axis  perpendicular  to  H,  and  one  inclined  45°  to  H. 
Take  a  line  for  the  common  element,  at  relative  distances  of  5  and  3  from 
the  axes,  bisecting  their  angle. 

696.  Show  the  surface  of  a  cow's  horn  arch,  which  has  one  opening  a  12-ft. 
semicircle  at  A(2,  0,  0)  and  the  other  an  8-ft.  semicircle  at  B(3,  0  —  1). 
Draw  the  projections  of  a  2-ft.  square  stone  on  the  interior  of  the  arch, 


CURVED  LINES  AND  SURFACES 


161 


its  center  being  at  C(2f  +  |  —  x).     (Hint:  Bound  the  block  with  four 

planes,  and  find  their  intersections  with  the  surface.)     Scale  i  in.  =  1  ft. 

697.  Design  a  conoidal  top-piece  for  the  hip  roof  shown  in  Fig.  100.     Make 


Conical 
Conoidal  Top  Piece.  Helical  Spring. 

FIG.   100. — Practical  warped  surfaces. 


Reducing  Hood. 
(Cows  Horn) 


all  three  roofs  inclined  4o°,  and  make  the  conoid  any  suitable  propor- 
tions.    Have  the  center  of  the  circular  directrix  at  the  point  of  inter- 
section of  the  three  roof  planes,  the  straight  directrix  directly  over 
11 


162  PRACTICAL  DESCRIPTIVE  GEOMETRY 

that  point,  and  the  plane  director  perpendicular  to  the  straight  direc- 
trix.    Obtain  the  roof  intersections,  and  develop  the  conoid. 

698.  Develop  the  surface  of  the  automobile  mud  guard  (cylindroid)  to  scale 
(Fig.  100). 

699.  Develop  the  surface  of  the  reducing  hood  (warped  cone)  shown  in  Fig. 
100.     Any  convenient  scale. 

700.  Develop  the  surface  of  the  reducing  hood  (cow's  horn)  shown  in  Fig. 
100.  Any  convenient  scale. 


DOUBLE -CURVED  SURFACES 

220.  A  double-curved  surface  is  generated  by  a  curve  following 
a  curve  in  such  a  way  as  to  generate  a  surface  that  is  neither 
single-curved    or   warped.     A    double-curved    surface    contains 
no  straight  lines.     It  may  be  either  double  convex,  like  a  sphere 
or  ellipsoid,  or  it  may  be  concavo-convex,  like  a  torus,  or  like 
the  turned  surfaces  of  numberless  articles. 

As  all  important  double-curved  surfaces  are  also  surfaces  of 
revolution,  they  will  be  considered  under  that  head. 

SURFACES  OF  REVOLUTION 

221.  Any  line  revolving  about  an  axis  generates  a  surface  of 
revolution. 

222.  Single -Curved  Surfaces  of  Revolution. — If  the  generatrix 
is  a  straight  line  and  in  the  plane  of  the  axis,  the  surface  will  be 
single-curved.     The   Cylinder  of  Revolution   and   the   Cone   of 
Revolution  are  the  only  examples. 

223.  Warped  Surfaces  of  Revolution. — The  only  example  in 
this  class  is  the  Hyperboloid  of   Revolution  of  One  Nappe,  see 
Art.  206. 

224.  Double-Curved  Surfaces  of  Revolution. — Any  plane  curve 
revolving  about  an  axis  generates  this  surface. 

Sphere. — Generated  by  a  circle  revolving  about  one  of  its 
diameters. 

Ellipsoids,  i.  Prolate  Spheroid. — Generated  by  revolving  an 
ellipse  about  its  major  axis.  Its  shape  resembles  a  foot  ball. 

2.  Oblate  Spheroid. — Generated  by  revolving  an  ellipse  about 
its  minor  axis.  Its  shape  resembles  a  door  knob. 

Paraboloids. — Generated  by  revolving  a  parabola  about  either 
axis.  When  revolved  about  the  transverse  axis,  it  generates  the 
surface  used  in  search-light  reflectors.  Why? 


CURVED  LINES  AND  SURFACES 


163 


Torus. — Generated  by  revolving  a  circle  about  an  axis  that  is 
not  a  diameter,  usually  completely  outside  the  circle.  In  this 
case  it  is  called  a  ring,  or  annular  torus.  As  a  matter  of  fact,  a 
torus  is  often  generated  by  other  curves  than  the  circle,  or  by  a 
combination  of  curves.  The  architectural  detail  is  of  this 
character.  In  this  work  the  word  torus  will  be  used  to  mean 
the  annular  torus. 

Hyperboloid  of  Revolution  of  Two  Nappes. — Generated  by 
revolving  a  hyperbola  about  its  transverse  axis.  It  is  not  often 
encountered  in  drafting. 

Many  surfaces  generated  by  combination  curves  will  be  found, 
some  with  names  and  some  without. 

225.  Meridian  and  Right  Sections. — Every  plane  perpendicular 
to  the  axis  cuts  a  surface  of  revolution  in  a  circle.     Hence,  we 
may  even  conceive  a  surface  to  be 

generated  by  a  circle  moving  in  the 
direction  of  the  axis,  and  changing  its 
diameter  according  to  some  law.  Any 
surface  turned  in  a  lathe  is  generated 
in  this  way.  (See  Fig.  101). 

Every  plane  through  the  axis, 
called  a  Meridian  Plane,  cuts  the  sur- 
face in  a  Meridian  Line,  which  has  the 
identical  form  of  the  line  used  to 
generate  the  surface. 

Any  two  surfaces  of  revolution 
having  a  common  axis  will  intersect 
in  a  circle  or  circles,  perpendicular  to 
the  axis.  When  the  axis  is  parallel 
to  the  plane  of  projection  these  inter- 
sections appear  as  straight  lines.  (See 
Fig.  101). 

226.  Representation. — Two  views  of  double-curved  surfaces  of 
revolution  are  all  that  are  necessary,  and  usually  one  is  sufficient, 
and  that  in  outline.     A  center  line  on  V,  perpendicular  to  GL,  is 
almost  always  drawn,  and  the  meridian  section  parallel  to  V 
constitutes  the  F-projection.     The  //-projection  is  merely  the 
circle  of  a  diameter  equal  to  the  largest  dimension  of  the  gener- 
atrix, with  as  many  other  circles,  visible  or  invisible,  as  may  be 
necessary  to  show  the  various  intersections  of  the  surface.     Fig. 
102  shows  a  torus,  which  will  fairly  represent  all  such  surfaces. 


FIG.  101. — Surface  of  revolution. 


164 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Any  view  of  a  sphere  is  a  circle,  so  both  projections  of  any  sphere 
will  be  circles. 

227.  Problem  82. — To  locate  a  point  on  any  Surface  of  Revo- 
lution. 

Let  it  be  required  to  locate  0  on  the  torus  in  Fig.  102,  assuming 
either  projection. 

Analysis. — Any  plane  parallel  to  H  will  cut  the  torus  in  a 
circle. 


FIG.  102. 

Construction,  assuming  the  F-projection,  o'. 

1.  Through  o'  pass  a  plane^  parallel  to  H.      It  will  cut  a  circle 
of  a  radius  equal  to  OP  (o\  p'). 

2.  Draw  the  ^/-projection  of  that  circle,  with  its  center  at  p. 

3.  Project  o'  to  o  on  the  circle. 

Note. — As  o'  could  project  to  two  different  points  on  the  circle,  it  is 
necessary  to  specify  "front  surface"  or  "rear  surface"  to  make  it  exact. 
If  O  were  nearer  the  center  there  would  be  four  possible  locations,  two  on  the 
inside  surface. 

Construction,  assuming  the  ^-projection,  o. 

1.  Through  o  draw  an  arc  of  a  circle,  its  radius  equal  to  op. 

2.  Project   o1    (where   the   arc   crosses   the   meridian   that  is 
parallel  to  V)  to  the  given  meridian  curve,  at  o'r 

3.  Draw  the  trace  of  the  horizontal  plane  through  o\. 

4.  Project  o  to  o'  on  this  trace. 

Note. — Two  points  are  possible  in  this  case,  so  "upper  surface"  and 
"lower  surface"  must  be  specified. 


CURVED  LINES  AND  SURFACES  165 

228.  Tangent  Planes. — As  in  other  surfaces,  a  tangent  plane 
to  a  double-curved  surface  is  the  locus  of  all  tangent  lines   at 
any  given  point,  each  tangent  to  one  of  the  infinite  number  of 
curves  of  the  surface  passing  through  that  point. 

The  tangent  plane  to  any  surface  of  revolution  is  perpendicular 
to  the  meridian  plane  containing  the  point  of  tangency. 

229.  Problem  83.— To  pass  a  Plane  Tangent  to  a  Sphere  at  a 
Point  on  the  Surface. 

Analysis. — Pass  a  plane  through  the  point  perpendicular  to 
the  radius. 

Let  the  student  make  the  construction. 

230.  To  Pass  a  Plane  Tangent  to  any  Double-Curved  Surface 
of  Revolution  at  a  given  Point  on  the  Surface. 

Let  the  given  point  be  O  on  the  surface  of  the  torus,  Fig.  102. 
Analysis. — 1.  Draw  a  tangent  to  the  right  section  through  the 
point. 

2.  Draw  a  tangent  to  the  meridian  section  at  the  point. 

3.  These  two  tangents  will  determine  the  plane. 
Construction. — 1.  Draw  the  tangent  OM  to  the  circle  through 

O,  and  locate  its  T-piercing  point  M. 

2.  Revolve  O  to  oto\,  so  that  the  meridian  plane  will  be  parallel 
to  V. 

3.  Draw  the  tangent  o^n^  to  the  meridian  curve,  in  revolved 
position,  and  locate  its  j?7-piercing  point  N. 

4.  Counter-revolve  N  to  its  original  position  n'n,  into  the  trace 
of  the  meridian  plane  through  0. 

5.  The  two  tangents  will  now  determine  the  tangent  plane, 
Tt  being  parallel  to  om,  and  Tt'  being  drawn  through  m'. 

231.  Problem  84. — To  pass  a  plane  tangent  to  a  sphere  through 
a  given  straight  line. 

Analysis. — Regard  the  sphere  as  being  enveloped  by  a  cylinder 
parallel  to  the  given  line,  and  pass  the  plane  tangent  to  the 
cylinder.  The  plane  will  be  tangent  also  to  the  sphere. 

Construction. — Let  AB  (Fig.  103)  be  the  given  line,  and  O  be 
the  center  of  the  sphere. 

1.  Pass  a  plane,  Q,  through  0  perpendicular  to  AB. 

2.  Find  the  point,  P,  in  which  AB  pierces  Q. 

3.  Revolve  P  and  O  about  the  #-trace  of  Q  into  H  at  pt  and 
olf  and  draw  a  circle  at  ol  of  the  same  radius  as  that  of  the 
sphere. 

4.  Draw  a  tangent  from  p1  to  this  circle. 


166 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


5.  Find  its  ^-piercing  point  m  (where  it  crosses  Qq). 

6.  Draw  the  traces  of  T,  the  required  tangent  plane,  through 
m  and  the  piercing  points  of  AB,  c  and  d'. 

Note  1. — This  problem  usually  has  two  solutions,  but  if  the  line  AB 
pierces  the  sphere,  the  problem  is  impossible. 

Note  2. — The  projections  of  the  cylinder  were  not  drawn,  as  it  is  merely 
an  imaginary  assistance  to  the  analysis.  It  is  not  even  necessary  to  imagine 
it,  as  the  normal  plane,  Q,  will  have  to  pass  through  the  center  of  the  sphere, 
although  it  does  not  have  to  be  perpendicular  to  AB.  This  is  convenient, 
however,  and  would  be  necessary  if  the  cylinder  idea  were  carried  out. 


FIG.  103. 

232.  Problem  85. — To  pass  a  Plane  through  a  given  Line 
Tangent  to  any  Surface  of  Revolution. 

This  problem  is  more  curious  than  practical. 
Analysis. — First  Case :  If  the  given  line  intersects  the  axis 
extended. 

1.  Draw  a  cone  tangent  to  the  given  surface,  with  the  apex 
at  the  intersection  of  the  given  line  and  the  axis. 

2.  Pass  a  plane  through  the  given  line  tangent  to  the  cone.     It 
will  be  a  tangent  also  to  the  surface  of  revolution,  as  required. 

Analysis. — Second  Case :  If  the  given  line  does  not  intersect 
the  axis. 

1.  Revolve  the  given  line  about  the  axis,  thus  generating  a 
hyperboloid  of  revolution. 


CURVED  LINES  AND  SURFACES 


167 


2.  Draw  a  tangent  line  to  both  surfaces. 

3.  Revolve  this  line,  generating  a  cone,  to  which  the  given 
line  is  tangent. 

4.  Pass  a  plane  through  the  given  line  tangent  to  the  cone. 
It  will  be  tangent  to  the  given  surface. 

Construction. — Let  it  be  required  to  draw  a  plane  T  through 
AB,  Fig.  104,  tangent  to  the  torus,  whose  axis  is  OP. 

1.  Revolve  AB  about  OP,  generating  the  hyperboloid,  part 
of  whose  outline  is  the  curve  c/  a/. 

2.  Draw  the  tangent,  d/  e/,  to  torus  and  hyperboloid. 


FIG.  104. 

3.  Revolve  the  meridian  plane  in  which  D  and  E,  the  points 
of  tangency,  are  located,  until  the  point  E  reaches  the  line  AB. 
ED  is  then  a  line  tangent  to  both  surfaces  intersecting  AB, 
which  is  an  element  of  the  hyperboloid,  and  therefore  tangent  to 
the  cone  generated  by  ED. 

4.  Pass  the  plane  T  through  AB  and  ED. 

233.  Problem  86.— To  obtain  the  Intersection  of  a  Plane 
with  any  Surface  of  Revolution. 

Analysis. — The  simplest  curve  that  can  be  cut  from  a  surface 
of  revolution  is  the  circle,  which  is  cut  by  a  plane  perpendicular 
to  the  axis.  Therefore  a  series  of  planes  perpendicular  to  the 
axis  will  cut  straight  lines  from  the  plane  and  circles  from  the 
given  surface.  The  curve  of  intersection  will  pass  through  the 


168 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


various  intersections  of  the  straight  lines  and  circles.  Fig.  105 
shows  both  projections  of  the  line  of  intersection  of  the  plane 
T  with  a  torus. 

The  vertices  (high  and  low  points)  of  the  curve  may  be  exactly 
located  by  projecting  the  torus  and  plane  on  a  plane  perpen- 
dicular to  the  H-tr&CQ  of  the  given  plane.  The  vertices  always 
lie  in  the  meridian  plane  which  is  perpendicular  to  the-  given 
plane. 

Note  1. — Any  plane  will  cut  a  circle  from  a  sphere,  because  any  diameter 
of  the  sphere  may  be  considered  its  axis. 

Note  2. — The  foregoing  method  is  often  useful  in  obtaining  the  inter- 
sections of  planes  with  ruled  surfaces  of  revolution,  particularly  the 
hyperboloid. 


FIG.  105. 

234.  Development  of  Double -Curved  Surfaces  of  Revolution.— 

Development  in  the  strict  sense  is  impossible,  because  there 
are  no  straight  lines.  However,  in  ornamental  sheet  metal 
work  and  the  manufacture  of  globes,  approximate  surfaces  are 
made  from  flat  sheets. 

This  is  done  by  two  methods,  the  "  Meridian  Method,"  in) 'which 


CURVED  LINES  AND  SURFACES 


169 


strips  are  cut  along  the  meridian  lines,  and  the  "  Zone  Method  " 
in  which  belts  of  the  surface,  zones,  are  regarded  as  conical 
surfaces.  Both  methods  are  shown  in  Fig.  106  and  should 
require  no  explanation. 


Meridian   Method. 


235. 


Zone  -Method. 
FIG.  106. — The  approximate  development  of  a  sphere. 

EXERCISES 


701.  Locate  A (2  +  £,  x)  on  the  front  surface  and  B(4,  y  — £)  on  the  lower 
surface  of  the  3-in.  sphere,  whose  center  is  C(3,  0,  0). 

702.  Locate  D(2|,  y-f)  on  the  lower  surface,  and  E(3f +  f,  z)  on  the  rear 
surface  of  the  1  J-in.  sphere,  whose  center  is  F(3  + 1  —  1). 

703.  Locate  G(3J,  x  — f)  on  the  upper  surface  of  the  sphere  in  Ex.  701,  and 
draw  the  traces  of  the  plane  tangent  at  that  point. 

704.  Locate  K(3J  +  £,  yLpn  the  rear  surface  of  the  sphere  in  Ex.  702,  and 
find  the  traces  of  the  plane  tangent  at  that  point. 

705.  Sphere  in  Ex.  701.     Find  the  section,  and  its  true  size,  made  by  the 
plane  T(l+2)  5(1-3). 

706.  Sphere  in  Ex.  702.     Find  the  section,  and  its  true  size,  made  by  the 
plane  S(  +  2)oo(- 1J). 


170  PRACTICAL  DESCRIPTIVE  GEOMETRY 

707.  Find  the  points  in  which  K(l,  0-2)  L(5  +  lf,  0)  pierces  the  sphere  in 
Ex.  701. 

708.  Find  the  points  in  which  M(2  +  l|-l)  N(4  +  $-|)  pierces  the  sphere 
in  Ex.  702. 

709.  Develop  the  surface  of  a  l$-in.  sphere  by  the  zone  method. 

710.  Develop  the  surface  of  a  l$-in.  sphere  by  the  meridian  method. 

711.  Draw  the  projections  of  an  oblate  spheroid,  its  minor  axis  A(3  +  $  —  If) 
B(3  +  2$-lf),    and   its    major   axis    C(l$  +  l$-lf)    D(4£  +  li~lf). 
Locate  the  points  E(2$  +  2,  x)  on  the  front  and  F(4,  y-1}.)  on  the 
lower  surface. 

712.  Find  the  traces  of  the  plane  tangent  to  the  oblate  spheroid  in  Ex.  711, 
atF(4,  y-li). 

713.  Oblate  spheroid  in  Ex.  711.     Find  the  traces  of  a  tangent  plane  con- 
taining G(44  +  1  -i>  ||  to  H  (1  +H-|)K(0,  0-1). 

714.  Oblate  spheroid  in  Ex.  711.     Find  the  section,  and  its  true  size,  made 
by  T(l  +  3)  5(3-3). 

715.  Oblate  spheroid  in  Ex.  711.     Find  where  it  is  pierced  by  M(l+3  — 2) 
N(5  +  l-l). 

716.  Develop  by  the  zone  method  the  spheroid  in  Ex.  711. 

717.  Draw  the  projections  of  the  prolate  spheroid,  whose  major  axis  is 
O(3  +  i-l$)  P(3  +  3i-l$),  and  whose  minor  axis  is  A(2  +  lf-l$) 
B(4  +  lf-l$).     Locate  C(2$  +  2|,  x)  on  the  front  and  D(3$,  y-1) 
on  the  lower  surface. 

718.  Spheroid  in  Ex.  717.     Draw  the  traces  of  a  plane  tangent  at  D(3$, 
y— 1)  on  the  lower  surface. 

719.  Spheroid  in  Ex.  717.     Draw  the  traces  of  a  tangent  plane  containing 
the  line  E(l$,  0-1)  F(2$  +  2-$). 

720.  Spheroid  in  Ex.  717.     Find  the  section,  and  its  tru.e  size,  made  By  the 
planeS(l+3)  4$(4$  -3). 

721.  Spheroid  in  Ex.  717.     Find  the  section,  and  its  true  size,  made  by  the 
plane  R(  +  2)oo(-3). 

722.  Spheroid  in  Ex.  717.     Develop  the  surface  by  the  meridian  method. 

723.  Draw  the  projections  of  a  torus,  generated  by  a  1-in.  circle  revolving 
about    M(3,     0  — If)    N(3  +  2  — If),    as     an  axis,    with    a  center  at 
O(2  +  l  — If).     Locate  K(2$,  x  — 1£)  on  the  inside  upper  surface  and 
P(3f +  f  —  y)  on  the  rear  surface. 

724.  Torus  in  Ex.  723.     Draw  the  traces  of  the  plane  tangent  at  P(3f  +  f  -  y) 
on  the  rear  surface. 

725.  Torus  in  Ex.   723.     Draw  the  traces  of  a  tangent  plane  containing 
A(3$  +  2,  0)  B(5,  0-1). 

726.  Find  the  section  of  the  torus  (Ex.  723)  made  by  the  plane  T(l+3) 
4(3  —  3$).     Also  its  true  size. 

727.  Find  the  section  of  the  torus  in  Ex.  723  made  by  the  profile  plane 
through  C(3i  +  l-l). 

728.  Find  the  section,  and  its  true  size,  of  the  toruf  in  Ex.  723,  made  by  the 
plane  R(+2)oo(-3). 

729.  Draw  the  projections  of  an  ovoid,  whose  axis  is  M(3  +  $  —  1$)  N(3  +  3 
—  1$),   whose  upper  radius  is  $  in.,  and  lower  radius  f/in.     Locate 

,  x)  on  the  front  surface. 


CURVED  LINES  AND  SURFACES 


171 


730.  Find  the  traces  of  the  plane  tangent  to  the  ovoid  in  Ex.  729  at  the 
point  O(3| +  |,  x)  on  the  rear  surface. 

731.  Find  the  section,  and  its  true  size,  of  the  ovoid  in  Ex.  729,  made  by 
the  plane  T(l+2)  5(3-3). 

732.  Find  the  plane  tangent  to  three  spheres,  all  resting  on  H,  whose  respec- 
tive centers  are  A(2  +  l-l),  B(3  +  i~li)  and  C(3J  +  f-i). 

733.  Develop  the  surface  of  the  ornamental  sheet  metal  surface  of  revolution, 
shown  in  Fig.  107.     Any  convenient  scale. 

Note. — One  strip,  or  meridian,  will  be  sufficient. 


k—  -  24"--     — - 


\J 

I 

FIG.  107. — Lawn  ornament. 


INTERSECTIONS  AND  DEVELOPMENTS  OF  ALL  SURFACES 

236.  General  Rule  for  Intersections. — Pass  a  series  of  surfaces 
through  the  two  intersecting  surfaces,  which  shall  cut  the  simplest 
possible  lines  from  them.     These  lines,  thus  cut,  intersect  each 
other  two  and  two,  and  by  joining  these  points,  we  obtain  the 
curve  of  intersection. 

Most  of  the  auxiliary  surfaces  used  for  this  purpose  are 
planes,  and  sometimes  spheres  are  used,  but  other  surfaces 
very  seldom,  never  in  the  writer's  experience. 

237.  The   Simplest  Lines   Cut  from  Surfaces  by  Planes.— In 
selecting  planes  to  cut  two  intersecting  surfaces,  the  following 
table  is  useful,  showing  how  to  pass  the  proper  planes  to  cut  the 
simplest  lines. 


172  PRACTICAL  DESCRIPTIVE  GEOMETRY 

Surface  Intersection  Kind  of  Plane 

Cone  Straight  line  Through  the  apex. 

Cone  Circle  Parallel  to  circular  base. 

Cylinder  Straight  line  Parallel  to  the  axis. 

Cylinder  Circle  Parallel  to  a  circular  base. 

Convolute  Straight  line  Tangent  to  helical  cylinder. 

Convolute  Involute  Perpendicular  to  the  axis. 

Helicoid  Straight  line  Through  the  axis. 

Warped  surface  .      Straight  line  According  to  its  generation. 

Surface  of  revolution  Circle  Perpendicular  to  the  axis. 

Surface  of  revolution  Meridian  curve  Through  the  axis. 

It  may  be  noted,  that  planes  cutting  straight  lines  from 
warped  surfaces  are  seldom  easy  to  use  because  of  the  variety 
of  their  directions.  There  are  cases  of  surfaces  having  a  plane 
director  where  it  is  convenient  to  get  straight  line  intersections. 

It  is  often  not  possible  to  use  the  planes  in  the  foregoing  list 
for  both  of  the  intersecting  surfaces,  but  it  is  always  possible 
to  use  them  for  one.  In  such  cases  the  second  surface  may  be 
cut  in  somewhat  complicated  curves.  It  is  such  cases  as  these 
that  make  it  impossible  to  lay  down  rules  for  all  cases.  The 
draftsman  must  use  his  judgment  and  instinct  for  the  right 
auxiliary  surfaces. 

CONCENTRIC  SPHERES  METHOD 

When  two  surfaces  of  revolution  have  intersecting  axes 
the  easiest  method  of  obtaining  their  intersection  is  to  use 
concentric  spheres.  By  employing  a  sphere,  as  an  auxiliary 
surface,  with  its  center  at  the  intersection  of  the  axes,  circles  are 
cut  from  the  two  surfaces,  because  the  intersections  of  two 
surfaces  of  revolution  having  a  common  axis  will  be  circles; 
Article  225.  A  typical  construction  will  be  shown  in  Problem  90. 

PROBLEMS  IN  INTERSECTIONS 

A  few  representative  problems  of  the  many  that  might  be  given 
are  selected  for  illustration. 

238.  Problem  87.  To  find  the  intersection  between  two  cones. 

Even  in  so  specific  a  problem  as  this,  the  position  and  character 
of  the  two  cones  have  much  to  do  with  the  choice  of  auxiliary 
planes  to  be  used.  We  give  two  representative  cases,  and  both 
could  be  successfully  worked  by  using  another  systeni  of  auxiliary 
planes. 


CURVED  LINES  AND  SURFACES 


173 


First  Case. — When  the  cones  are  oblique  and  their  bases  are 
circles  in  the  same  plane. 

Analysis. — Pass  a  series  of  planes  parallel  to  the  bases,  inter- 
secting the  cones  in  circles.  Study  Fig.  108  for  the  solution 
and  construction.  Let  the  student  work  this  by  passing  planes 
through  the^arjicesof  the  cones.  The  advantage  of  the  latter 
method  is  that  it  gives  definitely  the  outside  elements  of  the 
cones  that  are  involved  in  the  intersection. 


FIG.   108. — The  intersection  of  two  cones,  obtained  by  passing  planes  parallel  to  the  bases. 

239.  Second  Case. — When  the  cones  are  right  circular,  and 
their  bases  are  in  different  planes. 

Analysis. — 1.  Pass  a  series  of  planes  through  the  apices  of  the 
cones,  cutting  elements  from  each. 

2.  Connect  the  intersections  of  the  elements  in  order. 

Note. — In  these  and  succeeding  problems,  it  often  only  complicates  the 
problem  to  divide  each  or  either  surface  into  a  number  of  equal  parts.  The 
best  way  is  to  let  the  auxiliary  surfaces  determine  which  elements  to  draw. 

Construction. — To  find  the  intersection  between  the  two  right 
circular  cones,  whose  apices  are  A  and  B,  and  whose  bases  are 
in  P  and  H  respectively  (Fig.  109.) 


174 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


1.  Connect  A  and  B  with  a  line.     This  line  will  be  contained 
in  all  the  planes  passing  through  the  apices,  and  its  piercing 
points  will  lie  in  all  the  respective  traces  of  these  planes. 

2.  Find  the  ^-piercing  point,   C,   of  the  line  AB;   and  the 
P-piercing  point,  D,  of  the  same  line. 

3.  Through  c  draw  any  number  of  traces  of  auxiliary  planes, 
and  draw  the  elements  cut  by  them  from  the  cone  B. 


FIG.   109. — The  intersection  of  two  cones  obtained  by  passing  planes  through  their  apices. 

4.  Draw  the  profile  traces  of  the  same  auxiliary  planes,  and 
draw  the  elements  cut  by  them  from  the  cone  A. 

5.  Locate    the    intersections    of    all    the    elements    with    the 
elements  lying   in   their  respective  planes,  obtaining  the  H-,  V- 
and  P-projections  of  each. 

6.  Draw  the  line  of  intersection  through  them  in  order,  and 
determine  the  visible  and  invisible  parts  of  eaqh  projection  of 
the  curve. 


CURVED  LINES  AND  SURFACES 


175 


240.  Problem  88. — To  find  the  intersection  of  two  cylinders. 

This  problem,  like  those  preceding,  may  have  more  than  one 
means  of  solution,  according  to  their  relative  situations  and 
characteristics. 

In  the  case  of  the  cylinders  in  Fig.  110,  two  methods  are 
available,  as  follows: 

(A)  Pass  a  series  of  planes  parallel  to  both  axes,  cutting 
elements  from  both  surfaces. 


r  -q 


FIG.   110. — The  intersection  of  two  cylinders  found  by  passing  planes  parallel  to  both  axes. 

(B)  Pass  a  series  of  planes  parallel  to  both  circular  bases, 
cutting  circles  from  the  surfaces.  Method  B  can  only  be 
employed  when  the  circular  bases  are  in  the  same  plane,  as 
they  are  in  this  figure. 

In  other  cases,  it  may  be  necessary  to  pass  a  series  of  planes 
parallel  to  one  of  the  planes  of  projection;  or,  parallel  to  one 
axis  and  to  the  circular  base  of  the  other  cylinder. 

In  the  specimen  solution,  here  given,  planes  are  passed 
parallel  to  both  axes. 

Construction. — Let  it  be  required  to  find  the  intersection 
between  the  cylinders  in  Fig.  110,  whose  axes  are  AB  and  CD, 
both  having  circular  bases  in  H . 

1.  Draw  the  #-trace  of  a  plane  Q,  passed  through  AB  par- 
allel to  CD.  See  Problem  8. 


176  PRACTICAL  DESCRIPTIVE  GEOMETRY 

2.  This  trace  intersects  the  base  circles  at  m,  n,  o  and  p,  with 
F-projections  at  m',n',o'  and  p'  in  GL. 

3.  Through  M,  N,  O,   and  P  draw  elements  parallel  to  the 
respective  axes. 

4.  These  four  elements  lie  in  the  same  plane,  and  will  intersect 
each  other  in  four  points  on  the  curve.     Locate  these  points. 

5.  Draw  a  series  of  traces  parallel  to  qq,  and  repeat  operations 
(2),  (3),  and  (4)  with  each. 

6.  Draw  a  smooth  curve  through  all  of  the  points,  taking 
them  in  order. 

Note. — Do  not  divide  either  circle  into  any  number  of  equal  arcs.  Let 
the  planes  determine  the  elements. 

241.  Problem  89. — To  find  the  intersection  of  a  cone  and 
cylinder. 

The  following  methods  of  passing  planes  may  be  employed: 

1.  Through  the  apex  of  the  cone,  parallel  to  the  axis  of  the 
cylinder. 

2.  Parallel  to  the  bases  in  any  plane.     (This  method  is  very 
convenient  if  both  surfaces  have  circular  bases  in  H,  V  or  P.) 

3.  Parallel  to  the  elements  of  the  cylinder  and  perpendicular 
to  the  axis  of  the  cone,  if  it  be  a   right  circular  cone,  if  the 
conditions  are  right. 

Concentric  Sphere  Method. — If  both  are  surfaces  of  revolution 
with  axes  intersecting,  the  process  in  Problem  90  is  usually  the 
best. 

Auxiliary  Plane  Method. — There  are  certain  conditions  when 
the  easiest  and  neatest  solution  is  made  by  projecting  both  sur- 
faces on  a  plane  perpendicular  to  the  axis  of  the  cylinder.  The 
specimen  solution  given  here  is  for  such  a  condition. 

Construction. — Let  AB  (Fig.  Ill)  be  the  axis  of  a  right  cir- 
cular cone,  lying  on  H,  and  CD  be  the  axis  of  a  circular  cylinder, 
also  lying  on  H. 

1.  Draw  the  projections  of  the  cone  and  cylinder,  drawing  a 
sufficient  number  of  elements  of  the  cone. 

2.  Draw  the  #-trace  (qq)  of  an   auxiliary  plane  Q,  perpen- 
dicular to  the  axis  of  the  cylinder,  and  therefore  perpendicular 
toH. 

3.  Project  the  two  surfaces  on  Q.     The  cylinder  will  project 
as  a  circle,  tangent  to  qq,  and  the  elements  of  the  cone  will 
project  as  follows:  A  will  project  to  ax  on  qq,  as  a'  is  on  GL,  and 


CURVED  LINES  AND  SURFACES 


177 


the  points  1,  2,  3,  etc.,  will  project  the  same  distances  above  qq 
that  their  respective  V-projections  are  above  GL.     Why? 

4.  From  the  points  on  the  circle,  where  it  is  crossed  by  the 
various  elements,  project  to  the  various  elements  in  their  H- 
projections,  and  thence  to  their  F-projections. 

5.  Draw  the  curve  through  the  points  in  order. 


Hint. — To  avoid  confusion,  number  your  points  and  elements, 
because  the  following  put  of  the  correct  order  of  these  points  is 
often  very  difficult. 

Note. — This  problem  can  be  done  easily  also  by  passing  planes  through 
the  apex  of  the  cone,  parallel  to  the  elements  of  the  cylinder,  but  the  method 
given  is  preferable. 
12 


178 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


242.  Problem  90. — To  find  the  intersection  of  any  two  surfaces 
of  revolution  whose  axes  intersect. 

Analysis. — 1.  At  the  intersection  of  the  axes  as  a  center  draw 
the  projections  of  a  number  of  concentric  spheres. 

2.  These  spheres  will  cut  circles  from  the  two  surfaces;  see 
Art.     237. 

3.  The  intersections  of  the  proper  circles  will  yield  points  on 
the  required  intersection. 


6  ? 


FIG.  112. — The    employment  of  concentric  spheres  to  determine  the  line  of  intersection 
between  two  surfaces  of  revolution,  having  intersecting  axes. 

Construction. — Let  the  circular  cylinder,  whose  axis  is  A  B 
(Fig.  112),  and  the  ellipsoid,  whose  axis  is  CD,  be  the  two  surfaces 
of  revolution,  with  axes  intersecting  at  O. 

1.  With  o'  as  a  center,  draw  the  F-projections  of  several 
spheres.  The  y-projections  of  the  circles  cut  by  these  spheres 
will  be  straight  lines.  Why? 


CURVED  LINES  AND  SURFACES  179 

2.  The  circles  cut  from  the  ellipsoid  will  project  on  H  as  circles. 
Draw  these  projections. 

3.  From  the  F-projections  of  the  intersections  of  the  proper 
circles  project  to  the  circles  1,  2,  3,  etc.     To  do  this  it  will  not 
be  necessary  to  draw  the  /^-projections  of  the  cylinder  circles. 
Why? 

4.  Draw  the  two  projections  of  the  lines  of  intersection  through 
these  points  in  order. 

Note  1.  —  The  elements  of  the  cylinder  are  not  necessary  in  this  solution. 

Note  2.  —  If  the  position  of  the  surfaces  is  not  as  advantageous  as  in  this 
figure,  project  them  on  an  auxiliary  plane  parallel  to  both  axes.  However, 
nearly  all  such  problems  are  given  in  this  position,  and  the  difficult  positions 
will  hardly  ever  be  encountered. 

243.  Discussion  on  Intersections  and  Developments  of  Curved 
Surfaces. 

In  attempting  this  kind  of  work  a  perfect  knowledge  of  the 
characteristics  of  each  surface  is  indispensable.  One  must  recog- 
nize from  the  projections  of  the  surface  whether  it  is  single- 
curved,  double-curved,  warped,  or  plane.  Much  choice  of 
auxiliary  surface  will  exist  even  when  the  surfaces  are  recog- 
nized, but  the  well-equipped  student  should  by  this  time  have 
sufficient  judgment,  or  intuition,  to  choose  the  most  suit- 
able sections.  You  must  not  be  limited  to  one  way  of 
working  these  problems,  but  should  have  all  the  resources 
available  for  every  possible  contingency,  and  be  able  to  apply 
them  to  the  problem  in  hand. 

244.  EXERCISES  IN  INTERSECTIONS  AND  DEVELOPMENTS 

734.  Find  the  line  of  intersection  of  the  2-in.  cylinder,  A,  whose  axis  is 
C(3,  0-1$)  D(3  +  2-li),  and  the  l£-m.  cylinder,  B,  whose  axis  is 


735.  Develop  the  surface  of  cylinder  A  in  Ex.  734. 

736.  Find  the  line  of  intersection  of  the  2-in.  cylinder,  A,  whose  axis  is 
K(3,  0-li)  L(3  +  2-lJ),  and  the  1-in.  circular  cylinder,  B,  whose 
axis  is  M(li  +  i~li)  N(4i  +  2-l|). 

737.  Develop  the  surface  of  cylinder  A  in  Ex.  736. 

738.  Develop  the  surface  of  cylinder  B  in  Ex.  736. 

739.  Find  the  line  of  intersection  of  the  l|-in.  cylinder,  A,  whose  axis  is 
C(3,  0-1^)  D(3  +  2-li),  and  the  l£-m.  circular  cylinder,  B,  whose 


740.  Develop  the  surface  of  cylinder  A  in  Ex.  739. 

741.  Develop  the  surface  of  cylinder  B  in  Ex.  739. 


180  PRACTICAL  DESCRIPTIVE  GEOMETRY 

742.  Find  the  intersection  of  the  2-in.  cylinder,  A,  whose  axis  is  G(3,  0  —  1{) 
H(3  +  2-li)  and  the  li-in.  cylinder,  B,  whose  axis  is  K(l  +  l-l$) 
L(5  +  l-l$). 

743.  Develop  the  surface  of  cylinder  A  in  Ex.  742. 

744.  Develop  the  surface  of  cylinder  B  in  Ex.  742. 

745.  Find  the  intersection  of  the  l$-in.  circular  cylinder,  A,  whose  axis  is 
M(2,  0  —  1)  N(4  +  2  —  1)  and  the  l|-in.  circular  cylinder,  B,  whose  axis 
isO(2  +  2-H)P(4,  0-U). 

746.  Find  the  intersection  of  the  elliptical  cylinders,  A,  whose  base  is  a  1-in. 
circle  in  H,  axis  C(2,0-2)D(4  +  2-  1),  and  B,  whose    base  is  a  l$-in. 
circle  in  H,  and  whose  axis  is  E(2  +  2-l$)F(4,  0-1$). 

747.  Find  the  intersection  of  the   l$-in.   circular  cylinder,  whose  axis  is 
A(2,  0-1)  B(2  +  2-l)  and  the  elliptical  cylinder,  base  a  l$-in.  circle 
in  H,  axis  C(3$,  0-2)  D(l+2-l). 

748.  Find  the  intersection  of  the  1^-in.  circular  cylinder,  A,  whose  axis  is 
E(l  +  f  —  $)  F(4  +  f  —  2),  and  the  elliptical  cylinder,  B,  whose  axis  is 
G(2$,  0-1)  H(3  +  2-$),  and  base  a  U-m.  circle  in  H. 

749.  Find  the  intersection  of  two  elliptical  cylinders  whose  axes  are  K 
(1$,   0-2)    L(4  +  2-l)    and  M(2  +  l$,   0)   N(3$  +  $-2),   and   whose 
respective  bases  are  a  li-in.  circle  in  H,  and  a  l$-in.  circle  in  V. 

750.  Find  the  intersection  of  the  l$-in.  cylinder,  whose  axis  is  A(3,  0  —  1$) 
B(3  +  2  —  1$)  and  the  equilateral  triangular  prism  of  1-in.  sides  whose 
center  line  is  C(l$  +  l-$)  D(4  +  l-2). 

751.  Develop  the  surface  of  the  cylinder  in  Ex.  750. 

752.  Develop  the  surface  of  the  prism  in  Ex.  750. 

753.  Find  the  intersection  of  the  l$-in.  cylinder,  axis  E(3,  0-1)  F(3  +  2-  1) 
and  the  equilateral  triangular  prism  of  1-in.  sides,  whose  center  line  is 

" 


754.  Develop  the  surface  of  the  cylinder  in  Ex.  753. 

755.  Develop  the  surface  of  the  prism  in  Ex.  753. 

756.  Find  the  intersection  of  the  right  cone,  axis  A  (3,  0-l)B(3  +  2-l), 
base  a  If  -in.  circle  at  A,  and  the  1-in.  cylinder,  whose  axis  is  C  (2  +  1  —  1) 
D(4  +  l-l). 

757.  Develop  the  surface  of  the  cone  in  Ex.  756. 

758.  Develop  the  surface  of  the  cylinder  in  Ex.  756. 

759.  Find  the  intersection  of  the  right  cone,  whose  axis  is  E(3,  0—1$) 
F(3  +  2  —  1$),  base  a  2-in.  circle  at  E,  and  the  1-in.  cylinder,  whose 
axis  is  G(2  +  l-lf)  H(4  +  l-l£). 

760.  Develop  the  surface  of  the  cone  in  Ex.  759. 

761.  Develop  the  surface  of  the  cylinder  in  Ex.  759. 

762.  Find  the  intersection  of  the  right  circular  cone,  tangent  to  H,  whose 
axis  is  K(l,  0  —  2)  L(3  +  l  —  2),  and  the  circular  cylinder,  tangent  to 
PI,  whose  axis  is  M(l$  +  f-2)  N(2f  +  f-$). 

Hint.  —  Project  the  surfaces  on  an  auxiliary  plane  perpendicular  to  the 
axis  of  the  cylinder. 

763.  Find  the  intersection  of  the  right  cone,   whose  axis  is  A(3,  0—1$) 
B(3  +  2  —  1$),  and  base  a  2-in.  circle  in  H,  and  the  1-in.  circular  cylinder, 
whose  axis  is  C(l$,  0-lf)  D(4  +  l$-lf). 


CURVED  LINES  AND  SURFACES  181 

764.  Find  the  intersection  of  the  right  cone  in  Ex.  763,  and  the  f-in.  circular 
cylinder,  whose  axis  is  E(3$,  0-lf)  F(3$  +  2-lf). 

765.  Develop  the  cone  in  Ex.  763. 

766.  Develop  the  cylinder  in  Ex.  763. 

767.  Find  the  intersection  of  two  oblique  cones,  whose  axes  are  E(2,  0  —  2) 
F(3$  +  2$-$),  and  G(4,  0-1*)  H(2$  +  lf-£),  with  bases  l$-in.  and 
l£-in.  circles  at  E  and  G. 

768.  Find  the  intersection  of  two  oblique  cones,  whose  axes  are  A(2,  0—1$) 
B(3  +  2-l$)  and  0(3$,  0-1$)  D(2$  +  l$-l$),  with  bases  l$-in.  and 
1-in.  circles  at  A  and  C. 

769.  Develop  the  surface  of  the  cone  EF  in  Ex.  767. 

770.  Develop  the  surface  of  the  cone  GH  in  Ex.  767. 

771.  Develop  the  surface  of  the  cone  AB  in  Ex.  768. 

772.  Develop  the  surface  of  the  cone  CD  in  Ex.  768. 

773.  Find  the  intersection  of  two  right  circular  cones,   whose  axes  are 
K(3,  0-1)  L(3  +  2-l)  and  M(2  +  l-l)  N(4  +  l-l),  and  bases  are 
2-in.  and  1^-in.  circles  at  K  and  N. 

774.  Develop  the  surface  of  the  cone  KL  in  Ex.  773. 

775.  Develop  the  surface  of  the  cone  MN  in  Ex.  773. 

776.  Find    the  intersection  of    two    right    circular    cones,  whose  axes  are 
P(3,  0-1$)  O(3  +  2-l$)  andR(4,  0-1$)  Q(4 +  !$-!$),  whose  bases 
are  2-in.  circles  at  P  and  R. 

777.  Develop  the  surface  of  the  cone  PO  in  Ex.  776. 

778.  Develop  the  surface  of  the  cone  RQ  in  Ex.  776. 

779.  Find  the  intersection  of  the  cones  (right  circular),  whose  axes  are 
A(3,  0-1)  B(3  +  2-l)  andC(l$,  0-1)  D(4  +  f-l).     The  base  of  the 
first  is  a  2-in.  circle  at  A,  and  the  second  cone  is  tangent  to  H. 

780.  Develop  the  cone  AB  in  Ex.  779. 

781.  Find   the  intersection   of  the   right   circular   cones,    whose  axes   are 
E(3,  0-2)  F(3  +  2$-2)   and  G(3  +  $-2)   H(4$  +  l$-2),  and   whose 
bases  are  2-in.  and  1-in.  circles  at  E  and  G. 

782.  Develop  the  surface  of  both  cones  in  Ex.  781. 

783.  Make  frusta  of  the  two  cones  in  Ex.  781  by  passing  planes  perpendicu- 
lar to  the  axes,  1  in.  from  F  and  $  in.  from  the  apex  H.     Develop  the 
surfaces  of  both  (Coffee  Pot). 

784.  Find  the  intersection  of  the  oblique  cone,  whose  axis  is  K(2-f  2,  0) 
L(3$  +  l-2),  base  a  l$-in.  circle  at  K,  and  the  right  cone  M(3,  0-1) 
N(3  +  2$  —  1),  whose  base  is  a  If-in.  circle  at  M. 

785.  Find  the  intersection  of  two  oblique  cones,  having  the  same  base,  a 
2-in.  circle  at .0(3,  0-1$),  and  their  vertices  at  P(2  +  2-l)  and  Q 
(4  +  l$-l). 

786.  Find  the  intersection  of  the  oblique  cone,  whose  axis  is  A(2,  0  —  2) 
B(4  +  2  — 1),  and  base  a  1-in.  circle  at  A,  and  the  elliptical  cylinder, 
whose  axis  is  0(3$,  0-1$)  D(2$  +  2-l),  and  base  a  l$-in.  circle  at  C. 

787.  Develop  the  cone  in  Ex.  786. 

788.  Develop  the  cylinder  in  Ex.  786. 

789.  Find  the  intersection  of  the  right  circular  cone,  whose  axis  is   E(3, 
0-1$)  F(3  +  2  — 1$),  and  base  a  2-in.  circle  at  E,  and  the  elliptical 
cylinder,  whose  axis  is  G(2  +  2,  0)  H(3  +  l-2$),  base  a  1-in.  circle  at  G. 


182  PRACTICAL  DESCRIPTIVE  GEOMETRY 

790.  Develop  the  surface  of  the  cone  in  Ex.  789. 

791.  Develop  the  surface  of  the  cylinder  in  Ex.  789. 

792.  Find    the   intersection   of   the   right   circular  cone,  axis  K(3,  0  —  1^) 
L(3  +  2  —  1^),  and  2-in.  base,  and  a  triangular  prism  of  1-in.  sides,  whose 
center  line  is  M(2  +  l-$)  N(4  +  l—  2). 

793.  Develop  the  surface  of  the  cone  in  Ex.  792. 

794.  Develop  the  surface  of  the  prism  in  Ex.  792. 

795.  Find  the  intersection  of  the  oblique  cylinder,  whose  axis  is  A(2,  0  —  2) 
B(4  +  3  —  •£),  base  a  1  J-in.  circle  in  H,  and  the  2-in.  sphere,  whose  center 


796.  Find  the  intersection  of  the  3-in.  sphere,  center  at  D(3  +  l$  —  2)  and  a 
circular  cylinder  of  1^-in.  diameter,  whose  axis  is  £  in.  from  D,  parallel 
to  H,  and  inclined  30°  to  V. 

797.  Find  the  intersection  of  the  2-in.  sphere,  whose  center  is  M(3  +  1$  —  1£) 
and  the  1-in.  circular  cylinder,  whose  axis  is  N(3  +  2  —  2)O(3,  0  —  2). 

798.  Develop  the  surface  of  the  cylinder  in  Ex.  797. 

799.  Develop  the  surface  of  the  sphere  in  Ex.  797  by  the  zone  method. 

800.  A  torus  is  generated  by  a  1^-in.  circle,  center  at  A(2  +  1  —  If),  rotating 
about  B(3,  0-lf)  0(3  +  3  -If).     Find  its  intersection  with  the  U-in. 
circular  cylinder,  whose  axis  is  D(2,  0—  If)  E(2  +  2-lf). 

801.  Develop  the  surface  of  the  cylinder  in  Ex.  800. 

802.  Torus  in  Ex.  800.     Find  its  intersection  with  the  1^-in.  circular  cylin- 
der,   whose   axis  is  F(2£,   0—  If)    G(4  +  2  —  If).     (Concentric   sphere 
method.) 

803.  Develop  the  surface  of  the  cylinder  in  Ex.  802. 

804.  Find  the  intersection  of  an  oblique  cylinder,  whose  axis  is  K(2,  0  —  2) 
L(4  +  3  —  1),  base  a  1-in.  circle  in  H,  and  the  oblate  spheroid,  whose 
major  axis  is  M(li  +  l  —  1$)  N  (4$  +  !•—!£),  and  whose  minor  axis  is 
0(3,  0-li)  P(3  +  2-li). 

805.  Find  the  intersection  of  the  right  circular  cone,  of  a  2-in.  base,  whose 
axis  is  A(2,  0  —  2)  B(2  +  2J  —  2),  and  a  2&-in.  sphere,  whose  center  is 


806.  Find  the  intersection  of  an  oblique  cone,  axis  D(2,  0  -  2)  E(3$  +  2$  -  1), 
base  a  2-in.  circle  in  H,  and  the  2-in.  sphere,  whose  center  is  F(3  +  lj 

-H). 

807.  Torus  in  Ex.  800.     Find  its  intersection  with  the  cone  in  Ex.  805. 

808.  Torus  in  Ex.  800.     Find  its  intersection  with  the  cone  in  Ex.  806. 

809.  Develop  the  surface  of  the  cone  in  Ex.  805,  806,  807  or  808. 

PRACTICAL  EXERCISES  IN  VARIOUS  SURFACES 

810.  Lay  out  the  patterns  for  the  concrete-mixer  hopper,  shown  in  Fig.  113, 
to  scale.     Two  spaces. 

811.  Find  the  line  of  intersection  of  the  plane  and  curved  surfaces  of  the 
connecting  rod  end,  shown  in  Fig.   113.     Draw  three  projections  to 
scale,  one  space. 

812.  Find  the  lines  of  intersection  of  the  various  surfaces  of  the  octagonal 
tie  rod,  or  turnbuckle,  shown  in  Fig.  113,  to  scale.     One  space. 

813.  Draw  the  projections  of  a  2-in.  hexagonal  nut,  having  a  45°  conical 


CURVED  LINES  AND  SURFACES 

^ 


183 


Connecting  Rod  End. 


B'^^'W 

—  V3L         \% 


Hopper  of  Concrete  Mixer. 


Octagonal    Tie    Rod. 


Fairchild     Monoplane  Wing. 


Oil  .  Tank 


Sugar  Scoop. 


JL.    . 


K------  — 2/'(?- 


*s 

_*.._£_ 


Bleriot    Monoplane    Wing. 
FIG.   113. — Practical  exercises  involving  various  surfaces. 


184  PRACTICAL  DESCRIPTIVE/ GEOMETRY 

chamfer,  and  determine  the  exact  line  of  intersection.     Full  size,  one 
space, 

814.  Draw  the  projections  of  an  octagonal  dome,  like  the  one  shown  in 
Fig.  113,  to  scale,  and  lay  out  the  pattern  for  one  strip  of  the  surface. 
One  space. 

815.  Find  the  line  of  intersection  in  the  head  of  the  oil  tank  shown  in  Fig. 
113.     Make  the  flange,  at  the  seam,  5  in.  wide,  and  draw  the  projec- 
tions of  20  equally  spaced  cone-head  rivets.     Develop  one  sheet  (with 
the  hole  for  the  head)  of  the  body  of  the  tank.     Do  not  draw  the 
entire  length  of  the  body.     Any  scale,  two  spaces. 

816.  Lay  out  the  patterns  to  scale  for  the  sugar-scoop,  shown  in  Fig.  113. 
Two  spaces. 

817.  Develop  the  surface,  to  scale,  of  the  wing  of  the  Fairchild  monoplane, 
shown  in  Fig.  113.     What  is  the  character  of  the  surface?     Two  spaces. 

818.  Develop  the  surface,  to  scale,  of  the  wing  of  the  Bleriot  monoplane, 
shown  in  Fig.    113.     What  is  the  character  of  the  surface?      Two 
spaces. 

819.  A  grain  elevator  has  a  2-ft.  square  chute  running  downward  at  5°  from 
the  perpendicular.     It  is  proposed  to  run  an  off-chute  at  the  bottom 
at  45°  from  the  perpendicular,  12  in.  square.     Design  the  transition 
piece  between  the  two  chutes.     Lay  out  its  pattern  to  scale.     Two 
spaces. 

820.  Lay  out  the  patterns,  to  scale,  of  the  hopper  for  a  stone  crusher,  shown 
in  Fig.  114.     Two  spaces.     Note  that  two  surfaces  are  conoidal. 

821.  Lay  out  the  patterns,  to  scale,  of  the  dust  pan,  shown  in  Fig.  114. 
Two  spaces. 

822.  Draw  the  line  of  intersection,  and  lay  out  the  pattern  for  the  escape 
pipe  joint,  shown  in  Fig.  114.     One  space. 

823.  Design  a  steamship  ventilator  like  the  one  shown  in  Fig.  114.     Lay 
out  the  patterns  to  scale.     Two  spaces. 

824.  Find  the  line  of  intersection  of  the  ventilator  flue  and  its  escape  pipe, 
shown  in  Fig.  114.     Lay  out  the  patterns  to  scale.     Two  spaces. 

825.  Find  the  line  of  intersection  of  the  reducing  hood  (warped  cone)  and 
its  escape  pipe,  shown  in  Fig.  114.     Lay  out  the  patterns  to  scale. 
Two  spaces. 

826.  Lay  out  the  patterns  to  scale  for  the  transition  piece  for  the  fan  con- 
nection shown  in  Fig.  115.     Two  spaces. 

827.  Grain  Hopper  (Fig.  115).     Lay  out  the  patterns  to  scale  for  the  grain 
hopper,  noticing  that  two  of  the  sides  are  hyperbolic  paraboloids,  and 
not  planes.     Two  spaces. 

828.  Measuring   Hopper  for   Automatic   Filling    (Fig.  115).     Lay  out  the 
patterns  for  this  conoid  surface  to  scale.     Two  spaces. 

829.  Draft  Pipes  for  Ventilator  (Fig.  115).     All  pipes  are  4  in.,  except  the 
center  one,  which  is  8  in.     Find  all  the  lines  of  intersection  and  lay 
out  the  patterns  for  the  various  pieces  without  duplication.     Two 
spaces. 

830.  Bucket  for  Chain  Conveyor  (Fig.  115).     Lay  out  the  pattern  to  scale. 
One  space. 

Note. — This  surface  is  not  a  right  circular  cone;  what  is  it? 


CURVED  LINES  AND  SURFACES 


185 


831.  Motor  Boat  Funnel  (Fig.  115).  The  seam  between  the  two  cones  is  in 
a  plane  at  ri^ht  angles  to  the  center  line  of  the  larger  cone.  Lay  out 
the  patterns  to  scale.  One  space. 


Ventilator  Flue 
with  Escape  Pipe. 

FIG.   114. — Practical  exercises  involving  various  surfaces. 


832.  Coffee  Pot  with  Cylindrical  Spout  (Fig.  115) .  Find  the  line  of  intersec- 
tion between  body  and  spout,  and  lay  out  the  patterns  to  scale.  Two 
spaces. 


186 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


833.  Kettle  with  Conical  Spout  (Fig.  115).  Find  the  line  of  intersection 
between  body  and  spout,  and  lay  out  the  patterns  to  scale.  Two 
spaces. 


k—  8-—>\ 


Transition  Piece 
for  Fan  Connection. 


Draft  Pipes- 


Pouring    Pot 
for   Metals. 


FIG.   115. — Practical  exercises  involving  various  surfaces. 

834.  Cylindrical  Pouring  Pot  for  Metals   (Fig.  115).     Find  the  intersection 
of  spout  and  body,  and  lay  out  the  patterns  to  scale.     Two  spaces. 

835.  Conical  Pouring  Pot  (Fig.  115).     Find  the  line  of  intersection  of  body 
and  spout,  and  lay  out  the  patterns  to  scale.     Two  spaces. 


CURVED  LINES  AND  SURFACES 


187 


836.  The  center  line  of  the  discharge  ell  of  the  pump  connection,  shown  in 
Fig.  115,  makes  a  bend  through  90°  on  a  12-in.  radius.  The  outside 
diameters  are  everywhere  8  in.,  and  the  walls  are  f  in.  thick. 

(a)  Find  the  line  of  intersection  on  the  outside. 

(b)  Find  the  line  of  intersection  on  the  inside. 

(c)  What  is  the  surface  of  the  bent  pipe? 


Square  Bin  with 
Double-Bend  Pipe  Elbow. 


Passenger  Car  Ventilator. 


K 30 


Casing  and  Connections  for  Blower. 
FIG.  116. — Practical  exercises  in  surfaces. 

EXCERCISES  TO  OCCUPY  AN  ENTIRE  SHEET. 

837.  Make  full  drawings,  lay  out  all  the  patterns,  with  dimensions  in  full, 
listing  each  piece,  of  the  Ash  Sifter,  shown  in  Fig.  116.     To  scale. 


188 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


838.  Make  full  drawings  to  scale,  lay  out  all  the  patterns,  with  dimensions 
in  full,  listing  each  piece,  of  the  Square  Bin,  with  Double-bend  Pipe 
Elbow,  shown  in  Fig.  116. 

839.  Make  full  drawings  to  scale,  lay  out  all  the  patterns,  with  dimensions 
in  full,  listing  each  piece,  of  the  Casing  and  Connections  for  the  Blower, 
shown  in  Fig.  116. 

840.  Make  full  drawings  to  scale,  lay  out  the  patterns,  with  dimensions  in 
full,   listing  each  piece,   of  the  Passenger  Car   Ventilator,   shown  in 
Fig.  116. 

841.  Make  complete  drawings  to  scale  of  the  Air  Draft  Box  with  Draft 
and  Blast  Pipes,  as  shown  in  Fig.  117.     Lay  out  all  the  patterns  with 
dimensions  and  lists. 


12 


FIG.   117. — Air  draft  box  with  draft  and  blast  pipes. 


CHAPTER  IV 

PERSPECTIVE  AND  ISOMETRIC  PROJECTION 
PERSPECTIVE   PROJECTION 

245.  Discussion. — Early  in  this  work  (Art.  1-2)  Perspective 
Projection  was  denned  and  its  variance  from  Orthographic 
Projection  was  discussed.  It  was  shown  that  the  resemblance 
of  a  picture  to  the  image  formed  on  the  retina  of  the  eye  of  the 
observer  comes  from  the  fact  that,  as  an  object  recedes,  it  "looks" 
smaller.  This  is  so,  owing  to  the  diminishing  angle  between  the 
limiting  rays,  as  illustrated  in  Fig.  1,  p.  2.  This  gradual  diminu- 
tion of  an  object  in  recession  must  ultimately  result  in  its  vanish- 
ing, and  this  vanishing  of  receding  objects  is  the  basis  of  the 
science  of  perspective.  From  it  we  get  our  idea  of  distance  in  a 
picture,  and  the  proper  treatment  of  it  by  the  artist  or  drafts- 
man is  what  gives  the  picture  its  " life-like"  appearance. 

Standing  between  the  tracks  of  a  railroad  (parallel  lines),  and 
looking  in  their  direction,  we  seem  to  see  them  gradually  approach 
each  other  and  finally  to  meet  at  a  distant  point  on  the  level 
of  the  eye.  All  horizontal  lines  seem  to  vanish  at  this  level, 
and  the  fact  that  this  line  is  called  the  Horizon  shows  the  connec- 
tion. The  horizon  is  an  imaginary  line,  of  course. 

From  this  the  first  rule  of  perspective  is  obtained,  as  follows: 

All  parallel  horizontal  lines  meet  one  another  (that  is,  they 
vanish)  at  the  horizon.  All  horizontal  lines  terminate  at  the 
horizon  in  a  picture,  except  those  that  are  parallel  to  the  plane 
of  the  picture,  called  the  Picture  Plane.  These  lines  are  parallel 
to  the  horizon  in  the  picture. 

All  parallel  lines  in  space  are  converging  lines  in  the  perspective, 
and  their  vanishing  points  are  located  on  the  picture  plane.  The 
location  of  the  vanishing  points  of  the  various  systems  of  parallel 
lines  is  the  important  operation  in  perspective  projection,  and, 
having  located  them,  it  is  quite  easy  to  finish  the  picture. 

All  lines  parallel  to  the  picture  plane  vanish  at  infinity;  that  is, 
they  will  be  drawn  in  their  true  space  angles  in  the  picture. 
Hence,  all  vertical  lines  are  drawn  vertically  in  the  picture. 

189 


190 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


VANISHING  POINTS 

246.  All  the  parallel  lines  in  any  system  vanish  in  one  point, 
be  they  all  in  the  same  or  different  planes.     Therefore,  if  the  eye 
is  turned  in  the  direction  of  the  lines  in  any  system,  the  line  of 
sight  will   meet  them   all  in  the  vanishing  point.     Hence,   to 
locate  on  the  picture  plane  the  vanishing  point  for  any  system, 
look  in  the  direction  of.  the  lines  of  that  system,  holding  the 
picture  plane  in  its  proper   position,  and   the  vanishing   point 
of  the  system  will  be  the  point  in  which  the  line  of  sight  pierces 
the  picture  plane. 

247.  The  Construction  of  a  Perspective  from  an  Orthographic 
Projection. 

Fig.  118  shows  a  set  of  steps  in  third  angle  projection,  some- 
what lower  than  the  horizon.     Do  not  confuse  the  horizon  with 


e  m 


Horizon 


FIG.  118. — Orthographic  projection  of  an  object. 

GL,  as  they  are  not  used  in  precisely  the  same  manner. 
Notice  that  this  object  has  nothing  but  vertical  and  horizontal 
lines. 

There  are  two  ways  of  placing  the  object  for  a  perspective, 
called  respectively  One-Point  and  Two-Point  perspective.  In 
the  one-point,  called  so  because  there  is  but  one  vanishing  point 
for  the  principal  horizontal  lines,  the  object  is  placed  so  that 
one  system  of  its  horizontal  lines  is  parallel  to  the  picture 
plane;  that  is,  in  the  relation  of  the  object  in  Fig.  118  to  V. 
This  style  is  usually  employed  for  interiors  of  buildings,  stage 
scenery,  etc. 


PERSPECTIVE  AND  ISOMETRIC  PROJECTION 


191 


In  the  two-point  perspective  the  object  is  placed  at  an  angle 
to  the  picture  plane;  that  is,  "cornerwise."  This  is  the  way 
most  objects  appear  to  the  casual  observer,  and  is  the  style 
employed  in  drawing  exteriors  of  buildings,  machine  parts, 
articles  to  be  shown  in  catalogues,  etc.  Two-point  perspective 
will  be  employed  in  this  construction. 

First:  Draw  the  line  GL,  as  in  Fig.  119.  It  is  the  #-trace  of 
the  picture  plane.  Place  GL  well  up  on  the  drawing,  so  as  to 
allow  plenty  of  room  for  the  picture. 


GL 


(W 


M2^ 

ps 

FIG.  119. — The  placing  of  the  object  and  locating  the  vanishing  points. 


Second:  Draw  the  ^-projection  (Diagram)  of  the  object  at 
any  angle,  say  30°,  from  GL. 

Third:  Locate  the  Point  of  Sight  (PS)  at  some  convenient 
point,  preferably  in  front  of  the  center  of  the  object  and  as  far 
away  as  space  will  allow.  If  the  point  of  sight  is  too  near  the 
object,  the  picture  is  distorted. 

Fourth:  Group  the  parallel  lines  of  the  object  into  two  systems, 
A  and  B. 

Fifth:  Draw  parallels  from  the  point  of  sight  to  the  two 
systems.  They  will  pierce  the  picture  plane  at  the  vanishing 
points.  Thus  Va  and  V6  will  be  the  respective  vanishing  points 
for  the  two  systems. 

The  next  step  is  shown  in  Fig.  120.     Draw  lines  from  ps  to  the 


192 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


various  points,  a,  b,  c,  etc.,  of  the  object.     These  are  lines  of 
sight  and  pierce  the  picture  plane  at  al7  bx,  cu  etc.,  on  GL. 


GL 


\       I  I  If 

\  \<y    ,/ 
\\j .--' 

vps 

FIG    120  — Drawing  the  rays  from  the  points  of  the  object,  and  locating  the  points  in  which 
the  rays  pierce  the  picture  plane. 


FIG.   121. — Starting  the  picture,  drawing  the  horizon,  etc. 

The  first  step  in  drawing  the  picture  is  now  shown  in  Fig.  121. 
Draw  the  Horizon  line,  as  shown,  at  any  convenient  place,  where 
the  space  is  clear,  above  or  below  the  point  of  sight.  Project 


PERSPECTIVE  AND  ISOMETRIC  PROJECTION 


193 


the  two  vanishing  points  to  the  horizon.  Project  the  points 
ai>  b1;  ci>  etc.,  found  in  Fig.  120,  indefinitely  to  the  space  below 
the  horizon,  because  in  this  case  the  object  is  entirely  below  the 
horizon  (see  Fig.  118). 

All  points  of  the  object  that  are  in  the  picture  plane,  are  pro- 
jected on  the  picture  plane  in  their  true  distance  above  or  below 
the  horizon. 

Hence  the  points  E"  and  M  in  this  object  are  immediately 
located  at  e'  and  m',  their  true  distance  below  the  plane  of  the 
horizon. 


\\y\^      l  •  ILL lc 

FT  iFTTT    ID'S'TMTI  x 


FIG.  122. — The  picture  completed. 

These  points  alone  cannot  determine  all  of  the  points  in  the 
picture.  The  points,  F,  G,  H,  K  and  L,  all  lie  in  one  plane  and 
are  intersections  of  lines  of  both  A  and  B  systems,  and  verticals 
as  well.  Therefore,  if  we  obtain  the  picture  plane  trace  of  the 
plane,  P,  of  these  points,  lines  of  both  systems  may  be  drawn 
through  the  points  thus  determined. 

On  either  side  of  the  figure  draw  the  profile  view  of  the  steps, 
in  its  true  relation  to  the  plane  of  the  horizon.  Project  the 
points  f,  g,  h,  k  and  1  to  the  trace  p'p'.  Fig.  122  shows  these 
points  connected  with  Vb,  and  intersecting  the  projectors  in 
the  correct  points  on  the  picture  plane.  Connecting  these 
points  with  Va  will  be  sufficient  to  determine  the  lines  of  the  A- 
system.  A  little  study  of  Fig.  122  will  show  how  this  is  effected. 

13 


194 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


248.  The  Construction  of  an  Object  having  Oblique  Lines. 

Such  a  construction  as  the  picture  of  the  bungalow  shown  -in 
Fig.  123 ,  in  orthographic  projection,  involves  no  new  principles, 
merely  an  elaboration  of  the  principles  already  used  in  the 
previous  construction.  For  instance,  to  locate  on  the  picture 
plane  points  A  and  B  of  the  ridge: 

1.  Extend  AB  in  the  diagram  in  Fig.  124,  until  it  pierces  the 
picture  plane  at  C.  An  examination  of  the  projections  of  AB 


n 


Horizon 


Projections  of  a  Bungalow. 

FIG.  123. 

will  show  that  it  is  parallel  to  H  and  a  certain  distance  above 
the  horizon,  and  belongs  to  the  A-system  of  "parallels. 

2.  Having  located  the  point  c'  in  the  foregoing  operation, 
connect  it  with  Va.  The  points  a'  and  b'  are  located  on  this 
line  by  projecting  the  piercing  points  a/  and  b/  of  the  lines  of 
sight.  All  other  horizontal  lines  are  treated  in  the  same  manner, 
and  the  oblique  lines  are  drawn  from  point  to  point  on  the 
horizontals. 


249. 


EXERCISES  IN  PERSPECTIVE 


In  the  following  exercises  it  is  advisable  to  draw  the  projections  complete 
before  starting  the  perspective.  Place  the  object  for  each  perspective 
"  cornerwise,"  as  in  the  two  examples  just  given.  No  dimensions  are  neces- 
.sary  to  be  shown  in  these  pictures,  and  invisible  lines  should  not  be  drawn. 
The  pictures  should  be  drawn  as  large  as  space  will  permit,  but  care  should 


PERSPECTIVE  AND  ISOMETRIC  PROJECTION  195 


196 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


be  taken  to  have  the  point  of  sight  far  enough  from  the  picture  plane  (in 

proportion  to  the  size  of  the  object)   to  avoid  distortion.     Place  all  small 

objects  below  the  horizon. 

842.  Draw  the  perspective  of  a  l|-in.  cube. 

848.  Draw  the  perspective  of  the  frustum  of  a  square  pyramid,  lower  base 

1£  in.,  upper  base  f  in.,  altitude  (base  to  base)  1  in. 
844.  Draw  the  perspective  of  a  wooden  box,  to  scale,  20  in.  long,  12  in. 

wide,  6  in.  deep,  with  walls  1  in.  thick,  without  top. 


Core    Box 


Bearing   Box. 


FIG.  125. 


FIG.  126. 

845.  Draw  a  cross  made  of  five  1-in.  cubes  (like  Fig.  67),  surmounting  a 
3-in.  cube.     Draw  the  projections  and  perspective  to  scale,  with  the 
horizon  at  the  height  of  the  center  of  the  cross. 

846.  Place  a  1^-in.  equilateral  triangular    prism,   1£  in.  long  on  a   1-in. 


cube,  resembling  a  house  with  gable  roof, 
perspective. 


Make  the   drawing  and 


PERSPECTIVE  AND  ISOMETRIC  PROJECTION  197 

847.  Place  a  IJ-in.  square  pyramid,  J-in.  altitude,  on  a  1-in.  cube.     Make 
the  drawings  and  perspective. 

848.  Make  the  perspective  of  the  core  box  for  octagonal  core   shown  in 
Fig.  125. 

849.  Make  the  perspective  of  the  bearing  box  shown  in  Fig.  125. 

850.  Make  the  perspective  of  the  house  shown  in  Fig.  126.     Locate  the 
horizon  about  6  ft.  above  the  ground  line. 

ISOMETRIC  PROJECTION 

350.  Discussion. — A  convenient  and  very  satisfactory  mode  of 
making  working  drawings  on  a  single  plane  is  Isometric  Projec- 
tion, sometimes  called  Practical  Perspective.  Working  draw- 
ings, as  usually  made,  require  two,  and  often  three  or  more, 
views  to  present  all  the  necessary  details  and  measurements, 
but,  as  they  are  made  with  the  eye  at  an  infinite  distance,  they 
often  fail  to  convey  to  the  workman  a  correct  idea  of  the  actual 
appearance  of  the  object. 

By  making  the  drawing  in  Isometric  Projection,  we  imitate 
Perspective  in  placing  the  object  so  as  to  present  to  the  eye 
the  appearance  of  a  picture.  Yet  it  is  drawn  so  as  to  permit 
many  of  its  dimensions  being  made  to  scale.  But,  as  distance 
is  not  taken  into  account,  the  objects  have  a  more  or  less  dis- 
torted appearance  when  drawn  in  this  projection,  but  the  like- 
ness is  sufficient  for  its  purpose. 

The  advantage  of  Isometric  Projection  over  Perspective  is 
that  its  construction  is  very  much  easier  and  more  rapid.  It  is 
not  as  accurate  in  its  representation  as  perspective,  but  it  has 
the  advantage  of  being  made  to  scale,  as  it  actually  is  ortho- 
graphic projection  on  a  plane  placed  so  as  to  show  three 
dimensions. 

In  certain  lines  of  work,  such  as  the  piping  and  wiring  of 
buildings,  an  isometric  layout  is  very  desirable.  On  such  a 
layout  all  necessary  dimensions  can  be  shown,  and  the  con- 
tractor and  workmen  are  able  to  make  their  estimates  and  con- 
structions without  the  confusing,  and  often  uncertain,  labor  of 
using  a  multitude  of  blue  prints.  The  use  of  the  isometric 
layout  for  such  purposes  is  growing  into  general  adoption. 

251.  Definition. — Conceive  a  cube  placed  as  in  Fig.  127, 
presenting  its  corner  to  the  front.  Let  this  cube  be  projected 
on  a  plane  parallel  to  the  plane  of  the  points  A,  B  and  C.  The 
three  edges,  AO,  BO,  and  CO  will  be  at  equal  angles  to  the 
plane  of  projection,  and  will  be  projected  in  equal  lengths  and 


198 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


at  equal  angles  (120°)  to  each  other.  (Prove  this.)  The 
vertical  edge,  OB,  will  be  projected  as  a  vertical  line,  while  the 
horizontal  edges,  AO  and  CO,  will  be  projected  as  lines  making 
30°  with  the  horizontal.  The  complete  projection  of  the  cube 
will  be  as  shown  in  Fig.  128,  except  that  this  figure  is  actually 
the  projection  of  a  larger  cube,  although  its  edges  measure  the 
same  as  in  Fig.  127. 

252.  Isometric  Lines.— The  lines  AO,   BO   and   CO   will   be 
projected  in  equal  lengths  on  the  isometric  plane,  because  they 


a1 


crb 

Orthographic  Projection  of  a  Cube. 
FIG.  127. 


Isometric  Projection 
of  the  same  Cube. 


FIG.  128. 


are  equal  to  each  other  in  fact;  hence  the  name  "Isometric," 
which  means  equal  measurement.  These  three  lines  (or  any 
lines  parallel  to  them)  are  called  variously  directrices,  coordinate 
axes  and  isometric  lines.  We  shall  call  them  isometric  lines. 

Rule. — All  vertical  (isometric)  lines  are  drawn  vertically 
(as  in  Perspective),  and  all  horizontal  isometric  lines  are  drawn 
30°  from  the  horizontal,  in  one  direction  or  the  other,  according 
to  which  system  of  parallel  lines  they  are  members. 

Any  object  may  be  so  placed  as  to  give  an  isometric  view  by 
bringing  its  center  line  into  a  position  similar  to  this. 

253.  Isometric  Scale. — Although  the  isometric  lines  are  pro- 
jected in  equal  lengths  to  one  another,  their  projected  lengths 
are  not  as  long  as  their  true  lengths,  but,  .81647  of  the  true 


PERSPECTIVE  AND  ISOMETRIC  PROJECTION  199 

length.  In  Figs.  127  and  128,  the  cube  shown  in  orthographic 
and  isometric  projection  is  given  edges  of  equal  length  in  both 
figures.  Strictly  speaking,  this  should  not  be  done,  but  the 
lines  in  Fig.  128  should  all  be  .81647  of  the  length  they  are 
drawn.  Practically,  however,  no  attention  is  paid  to  this  fore- 
shortening, because  nearly  all  drawings  are  made  to  a  reduced 
scale,  so  that  it  is  not  necessary  to  have  an  agreement  in  scales 
between  the  two.  The  great  practical  feature  of  this  is  that 
dimensions  may  be  laid  off  on  isometric  lines,  using  any  con- 
venient scale.  This  means  that  isometric  drawings  are  made 
on  a  slightly  larger  (22%)  scale  than  the  orthographic  drawings. 
The  following  rules  will  be  found  useful  in  laying  out  isometric 
drawings. 

254.  Rules. — 1.  All  vertical  lines  in  the  object  are  vertical 
in  the  drawing,  and  are  drawn  in  their  true  length. 

2.  All  horizontal  lines,  parallel  to  either  Isometric  Horizontal, 
will  be  drawn  in  their  true  length,  inclined  30°  to  the  horizontal. 

3.  All  other  lines  will  be  foreshortened  or  lengthened. 

4.  Any  point  of  an  object  may  be  located  isometrically  by 
means  of  isometric  ordinates  from  known  lines. 

5.  All  straight  lines  may  be  located  isometrically  by  the  iso- 
metric projections  of  two  of  their  points. 

6.  All  curved  lines  may  be  located  isometrically  by  the  iso- 
metric projections  of  a  sufficient  number  of  their  points. 

•  255.  Problems  in  Isometric  Projection. 
Problem  91. — To  Draw  a  Cube. 
Note. — All  lines  in  the  cube  are  Isometric  lines. 

Construction. — Let  the  projections  of  the  cube  be  those  shown 
in  Fig.  127. 

1.  Draw  the  vertical  line  BO  (Fig.  128),  equal  in  length  to 
o'b'  (Fig.  127). 

2.  From  O  draw  two  30°  lines,  OA  and  OC,  equal  in  length 
to  BO. 

3.  Complete   the   three   parallelograms   determined   by  these 
three  lines  in  pairs. 

The  result  will  be  the  isometric  drawing  of  the  given  cube,  as 
shown  in  Fig.  128. 

256.  Problem  92.— To  Locate  a  Point. 

Construction.— Let  the  point  M  inside  the  cube  (Fig.  127)  be 
given;  to  find  its  location  inside  the  cube  drawn  in  Fig.  128. 

1    Assume  the  projections  m,m'  in  Fig.  127. 


200 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


2.  Draw  the  line  mn,  m'n'  parallel  to  OC,  piercing  the  face 
at  nn'. 

3.  From  N  draw  the  line  np,  n'p',  parallel  to  OB,  until  it 
intersects  the  edge  BD. 

These  lines,  MN  and  NP,  are  isometric  ordinates,  being  par- 
allel to  isometric  lines. 

4.  Lay  off  BP  on  BD  (Fig.  128)  equal  to  bp  (Fig.  127). 

5.  Lay  off  PN  parallel  to  BO  (Fig.  128)  equal  top'n'  (Fig.  127). 

6.  Lay  off  NM  parallel  to  OC  (Fig.  128)  equal  to  nm  (Fig.  127). 
M  will  be  the  required  point  inside  of  the  cube.     Any  point  in 

space  can  thus  be  located  on  an  isometric  drawing. 
257.  Problem  93.— To  draw  an  exact  circle. 


FIG.   129. 


For  this  and  the  two  following  problems,  let  us  use  the  cube, 
shown  in  orthographic  projection  in  Fig.  129,  and  in  isometric 
projection  in  Fig.  129a. 

Construction. — 1.  Inscribe  the  circle  in  the  square  B  of  the 
cube  in  Fig.  129. 

2.  In  each   corner   of  the   square   construct   a  small   square 
with  one  corner  in  the  circle  at  E,  G,  H  and  K;  example  CDEF. 

3.  Draw    these    small    squares   isometrically    in   Fig.    129a; 
with  corners  at  E',  G',  H',  and  K' ;  example  C'  D'  E'  F'. 

4.  Through  E',  G',  H',  and  K'  draw  a  smooth  ellipse  tangent 
to  the  four  sides  of  the  square  at  their  middle  points. 

258.  Problem  94. — To  draw  an  approximate  circle. 

This  is  the  system  almost  exclusively  used  in  actual  drafting, 
because  it  can  be  drawn  with  the  compasses. 

Construction. — Let  the  circle  be  drawn  in  face  A,  the  upper 
horizontal  face,  of  the  cube  in  Fig.  129. 


PERSPECTIVE  AND  ISOMETRIC  PROJECTION 


201 


1.  In  the   upper   parallelogram,  F'TO'S'  (Fig.   129a,)    draw 
lines  from  S'  and  T'  to  the  centers  of  the  opposite  sides,  inter- 
secting at  Q'  and  R'. 

2.  From  T'  as  a  center,  with  radius  TV  draw  the  arc  U'V, 
and  do  the  same  from  S'  as  a  center. 

3.  From  Q'  and  R',  as  centers,  draw  arcs  with  Q'U'  and  R'V 
as  radii,  completing  the  curve. 

259.  Problem  95. — To  draw  lines  making  any  required  angles 
with  any  given  line. 

Construction. — Let  it  be  required  to  draw  lines  making 
30°,  45°  and  75°  with  LM,  on  the  profile  face  of  the  cube. 

1.  Draw  LN,  LO  and  LP  in  the  projection  of  the  cube 
(Fig.  129). 


br  —  - 


B< 


>0 


FIG.  130a. 


2.  From  M'  lay  off  the  distance  MN,  on  M'O',  and  draw  L'X'. 
This  will  be  the  30°  line. 

3.  From  M'  lay  off  the  distance  MO,  to  the  point  0',  and 
draw  I/O'.     This  will  be  the  45°  line. 

4.  From  T  lay  off  the  distance  TP,  on  TO',  and  draw  L'P'. 
This  will  be  the  75°  line. 

Note. — If  the  angles  were  laid  off  at  M  instead  of  L,  they  would  work  out 
differently.  For  instance,  the  angles  at  I/  and  M'  each  represent  90°,  yet 
one  is  actually  60°  and  the  other  120°,  as  they  are  drawn  on  the  paper. 


202 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


260.  Problem  96. — To  draw  a  frustum  of  a  hexagonal  pyramid. 
Construction. — Let  it  be  required  to  draw  the  frustum,  whose 
projections  are  shown  in  Fig.  130. 


Pulley  24 


Hexagonal 
Bolt. 


Pipe    Tee. 


Corners  UK. 


i  K--2---H  I 
K--;3---H 
Bearing  Brass. 


K- 12"- H 

M  ission  Book  Rack. 


Crank  Forging 


^  .  __:  _____  ________  /2/_.  __________  ^ 

Pillow  Block. 
FIG.  131. 


1.  Circumscribe  about  each  base  a  rectangle.     The  sides  of 
these  rectangles  will  be  isometric  lines. 

2.  Draw  isometrically  (Fig.  130a)  the  rectangle  A'B'C'D'. 


PERSPECTIVE  AND  ISOMETRIC  PROJECTION  203 

3.  Locate  the  point  O'  by  ordinates. 

4.  Through   the   point   O'   draw   isometrically   the   rectangle 
E'F'G'H'. 

5.  Locate  by  ordinates  the  points  1,  2,  3,  4,  5,  6,  7,  and  8. 

6.  Connect  the  proper  points  in  order. 

Note. — Do  not  draw  in  the  invisible  lines.  They  are  drawn  here  to  make 
the  figure  complete,  but  in  practical  work,  where  isometric  drawings  are 
used,  the  dotted  lines  are  seldom  drawn. 

261.  Other  Methods  of  Pictorial  Projection. 

Under  certain  conditions,  a  closer  approximation  of  a  per- 
spective than  that  shown  by  the  isometric  is  obtained  by  the  use 
of  other  oblique  projections,  called  Cavalier  and  Cabinet 
Projections. 

Cavalier  Projection. — The  object  is  placed  with  one  face 
parallel  to  the  picture  plane,  and  the  projectors  are  considered 
inclined  45°  to  the  picture  plane.  Thus  one  face  is  the  same  as 
the  orthographic  view  would  be,  and  all  lines  perpendicular  to 
the  picture  plane  are  projected  in  their  true  length.  It  thus 
becomes  a  form  of  isometric  projection,  with  less  distortion, 
because  one  face  is  the  same  as  orthographic  projection  would 
make  it. 

Cabinet  Projection. — To  further  reduce  the  distortion,  the 
receding  lines  are  shortened  one-half.  Otherwise  the  drawing- 
is  made  like  cavalier  projection. 

Good  treatment  of  these  and  other  styles  of  single  plane  pro- 
jection, with  numerous  illustrations,  is  found  in  Prof.  French's 
"  Engineering  Drawing." 

EXERCISES  IN  ISOMETRIC  DRAWING 

262.  Make  isometric  drawings  of  the  following  objects,  to  scale  wherever 
necessary.     One  space  to  each. 

851.  Wooden  box,  without  cover,  10  in.  long,  6  in.  wide,  4  in.  deep,  made  of 
f-in.  boards.     Stand  it  on  end,  open  to  the  front. 

852.  Draw  the  box  in  Ex.  851  in  natural  position  with  a  cover,  open  120°. 

853.  Pentagonal  plinth,  2-in.  diameter,  \  in.  thick,  surmounted  by  a  1-in. 
cylinder,  \\  in.  high. 

854.  Bolt  shown  in  Fig.  131. 

855.  Book  rack  in  Fig.  131. 

856.  Pulley  24-in.  diameter,  8-in.  face,  2-in.  bore  (Fig.  131). 

857.  Bearing  Brass  for  2-in.  shaft  (Fig.  131). 

858.  Crank  Forging,  4-in.  throw  (Fig.  131). 
S59.  H-in.  Pipe  Tee  (Fig.  131). 


204 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


860.  Pillow  Block  for  2-in.  shaft  (Fig.  131). 

861.  Piping  Layout,  shown  in  Fig.  132.     This  is  intended  for  a  full  sheet, 
and  may  be  drawn  with  a  Title  and  Bill  of  Material.     A  tracing  made 
of  this  exercise  makes  a  splendid  drawing. 

Note. — First  make  the  layout  of  the  center  lines,  making  allowances  for 
all  the  connections. 


|x  6  "Pipe 


I  Elbow 


i«   o7' /^"7-_-''u:     ^±1  b 


2x1$.  Tee-' 

2"P/ug-*  ' 


FIG.  132.— Piping  layout 


INDEX 


Angle  between  a  line  and  plane,  64 

of  projection,  68 
Angle  an  oblique  plane  and  a  plane 

of  projection,  74 
Angle  two  planes,  72 
Angle  two  straight  lines,  37 
Angles  of  space,  4,  7 
Auxiliary  planes  of  projection,  83 
"Auxiliary  plane"   method    of    ob- 
taining   intersections,    83, 
176 

Axis,  definition,  109 
major,  109 
minor,  109 

B. 

Bisector  of  angles,  41 

C. 

Cabinet  projection,  203 
Cavalier  projection,  203 
Circle,  101 

definition,  107 

generating,  109 

pitch,  109 

rectification,  106 
Classification  of  lines,  101 

of  surfaces,  120 
Common  perpendicular  to  two  lines, 

81 

"Concentric    spheres"    method    of 
obtaining  intersections,  172, 
176,  178 
Cone,  120 

definition,  122 

development,  128 

exercises,  130 

of  revolution,  162 

problems,  123 

properties,  69,  122 

varieties,  122 


Conic  sections,  101,  107         ^ 
Conoid,  definition,  120 

discussion,  156 
Convolute,  120 

definition,  144 
development,  146 
exercises,  148 
helical,  144 
problems,  145 
Counter-revolution,  45 

by  similar  triangles,  46 
Cow's  horn,  120 

discussion,  156 
Curved  lines,  101 

classification,  101 
conic  sections,  101,107 
double,  101 
exercises,  119 
gearing  curves,  101,  107 
normals  to,  105 
plane,  101,  102 
projections  of,  102 
rectification  of,  105 
single,  101 
space  curves,  101 
tangents  to,  103 
Cycloid,  construction,  113 

definition,  107 
Cycloidal  curves,  101 
Cylinder,  120 

definition,  133 
development,  134 
exercises,  140 
projecting,  120 
of  revolution,  162 
Cylindroid,  120 

discussion,  156 

D. 

Designation  of  points,  7 
Development  of  cones,  128 

of  convolutes,  146 

of  cvlinders,  140 


205 


206 


INDEX 


Development  of  double-curved  sur- 
faces of  revolution,  168 

of  spheres,  169 

of  surfaces,  121 

of  warped  surfaces,  168 
Distance  from  a  point  to  a  line,  43 

from  a  point  to  a  plane,  63 
Directrix,  definition,  109 

of  a  parabola,  112 
Double-curved  surface,  162 
Double-ruled  surfaces,  149 

E. 

Element,  definition,  101 

of  contact,  125,  133 

of  helix,  118 

of  any  surface,  120 
Ellipse,  101 

applications,  110 

construction,  109 

definition,  107 

major  axis,  109 

minor  axis,  109 
Ellipsoid,  120 

definition,  162 
Epicycloid,  construction,  115 

definition,  108 
Evolute,  109 

F. 

Flat  representation,  3 

Focus,  109 

Frustum  of  a  cone,  122 

G. 

Gearing  curves,  107 
Generatrix,  120 
Gorge  circle,  153 

-H. 

Helicoid,  120 

approximate  development,  156 

"developable,"  144 

discussion,  151 
Helicoidal  arch,  152 
Helix,  101 

angle  of,  117 

conical,  133,  160 


Helix,  construction,  116 
definition,  108 
tangent  to,  118 
Horizon,  189 
Hyperbola,  101 

application,  112,  154 
construction,  111 
definition,  107 
Hyperbolic  paraboloid,  120 
definition,  154 
representation,  155 
second  generation,  155 
Hyperboloid    of   revolution    of    one 

nappe,  120 
of    revolution    of    one    nappe, 

appli  cation,  154 
of    revolution    of    one    nappe, 

discussion,  152 
of    revolution    of    one    nappe, 

representation,  153 
of    revolution    of   two    nappes, 

121,  163 

Hypocycloid,  construction,  115 
definition,  108 

I. 

Intersection   of  cone   and   cylinder, 

176 

of  two  cones,  172 
of  two  cylinders,  175 
Intersections  and  developments  of 

all  surfaces,  171 

and  developments,  exercises,  179 
and     developments,      practical 

exercises,  182 
and    developments,    problems, 

172 
obtained  by  concentric  spheres, 

172,  176 

by    auxiliary    plane    projec- 
tion, 83,  176 
Involute,  101 

construction,  116 
definition,  108 
Isometric  lines,  198 
projection,  197 
exercises,  203 
of  angles,  201 
of  a  circle,  200 


INDEX 


207 


Isometric  projection,  of   solids  hav- 
ing oblique  edges,  202 
problems,  199 
rules,  199 
scale,  198 

L. 

Layout  for  plates,  9 
Line  of  intersection  of  two  planes,  64 
of  shade,  94 
of  sight,  191 
Lines  perpendicular  to  lines,  16 

to  planes,  17,  58 
Location  of  points  in  space,  7 

M. 

Major  axis,  109 
Meridian  line,  163 

method  of  development,  168 
of  hyperboloid,  154 
plane,  154 
section,  163 
Minor  axis,  109 

N. 

Nappe  of  a  cone,  122 
Normal  line,  105 
Notation,  8 

O. 

Oblate  spheroid,  121,  162 
Oblique  section  of  a  surface  of  revo- 
lution, 167 

Orthographic  projection,  1 
P. 

Parabola,  100,  107 

construction,  112 
Paraboloid,  120,  162 
Perspective  projection,  1,  189 
construction,  190 
exercises,  197 
Pictorial  projection,  203 
Picture  plane,  189 
Piercing  points  of  lines,  18,  21,  22 

with  oblique  planes,  55 
Plane  figures  in  any  position,  44 
exercises,  48 
making    required    angles    with 

H  or  V,  78 

making    required    angles    with 
H  and  F,  68 


Planes,  notation,  15 
of  projection,  3 
parallel,  15,70 

parallel  to  lines,  28 
perpendicular,  15 
pierced  by  a  line,  55 
projecting,  15 
representation,  13 
theorems,  14 
Point  of  sight,  190 
Points  in  planes,  25 
notation,  12 
representation,  5 
revolution  of,  29,  32 
Practical  exercises,  28,  37,  43,  50,  60, 
65-67,  70,  72,  76,   77,  80, 
81,  83,  86-93,  100 
in  cones,  131 
in  cylinders,  142 
in  double-curved  surf  aces,  171 
in  isometric  projection,  203-4 
in  perspective,  196-7 
in  various  surfaces,  182-8 
in  warped  surfaces,  160-1 
Profile  plane,  3,  14 
projection,  11 
Projecting  cylinder,  102 

plane,  15 
Projection  of  a  line  on  an  oblique 

plane,  63 
Projections,  1 
Projectors,  9 

Prolate  spheroid,  121,  162 
Properties  of  the  right  circular  cone, 
69 

R. 

Rectification  of  arcs  of  circles,  106 

of  curves,  105 

of  quadrants,  106 
Representation  of  curves,  102 

of  planes,  13 

of  points,  5 

of  straight  lines,  13 
Revolution  of  lines,  32 

of  points,  29,  32 
Right  circular  cone,  properties,  69 

section,  163 
Rolling  hyperboloids,  154 


208 


INDEX 


Ruled  surfaces,  120 

S. 

Shade,  94 

Shades  and  shadows,  94 

exercises,  99 
Shadow  of  a  line,  95 
of  a  point,  95 
of  a  plane  figure,  96 
of  a  solid,  97 

Shortest  path  on  a  surface,  129 
Simplest  lines  cut  from  surfaces  by 

planes,  171 

Single  plane  projection,  189 
Solids,  exercises,  61 

to  draw  in  oblique  position,  58 
Surfaces,  119 

classification,  119 
double-curved,  120 
single-curved,  119 
warped,  120 
Surfaces  of  revolution,  120 

classification,  120,  162 
definition,  162 
hyperboloid,  152,  162 
problems,  164 
representation,  163 
ruled,  120 
single-curved,  162 
warped,  162 
Sphere,  120,  162 

development,  169 
tangent  planes,  165 
Spiral  of  Archimedes,  133,  151 
Spirals,  101 
Straight  lines,  10 
intersecting,  17 
lying  in  planes,  18,  25 
measurement  of,  33 

of  angles,  37 
parallel,  16 

to  H  or  V,  16 
perpendicular,  16 

to  a  plane,  17 
representation,  16 
revolution,  32 


T. 

Tangent,  construction,  104 
definition,  103 

planes    to    double-curved    sur- 
faces, 165 
planes  to  single-curved  surfaces, 

148 
planes  to  spheres,  165 

planes  to  warped  surfaces,  121 
rules,  148 

Torus,  120,  163,  164 
Triangulation,  123,  128,  150,  157 

U. 

Umbra,  94 

Uses  for  ellipse,  110 
for  hyperbola,  112 
for  parabola,  113 
for  warped  surfaces,  149 

V. 

Vanishing  points,  190 
Vertex,  109 

of  curve  of  intersection,  168  * 

W. 

Warped  cone,  120,  156 
quadrilateral,  155 
surfaces,  120 

development,  150,  156 

discussion,  149 

double  ruled,  150 

exercises,  158 

methods  of  generation,   149 

of  revolution,  162 

practical  exercises,  162 

representation,  150 
Weight  of  lines  for  exercises,  19 

Z. 

Zone  method  of  developing  double- 
curved  surfaces  of  revolu- 
tion, 169 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 
BERKELEY 

Return  to  desk  from  which  borrowed. 
This  book  is  DUE  on  the  last  date  stamped  below. 


RECTD  LD 

MAR    61961 


LD  21-100w-ll,'49(B7146sl6)476 


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UNIVERSITY  OF  CAUFORN1A  LIBRARY 


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